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A skew Young diagram is connected when any two of its cells can be joined by a sequence of cells sharing edges; touching only at a corner does not connect them. A nonempty border strip, or rim hook, is a connected skew diagram containing no $2\times 2$ square. Its height is its number of occupied rows minus one.

For a weak composition $\alpha =({\alpha }_1,\dots ,{\alpha }_{\ell })$, a border-strip tableau of shape $\lambda /\mu$ and type $\alpha$ is a chainwhose $i$th difference is a border strip of size ${\alpha }_i$. Filling that strip with $i$ gives the equivalent weakly increasing row-and-column filling. A zero part corresponds to an empty difference and has height zero. Put $\mathrm{ht}(T)={\sum }_i\mathrm{ht}({\lambda }^{(i)}/{\lambda }^{(i-1)})$. For positive $r$, $p_r={\sum }_j{x}_j^r$; write $p_{\alpha }={\prod }_{{\alpha }_i>0}{p}_{{\alpha }_i}$. Thus zero type parts are ignored: one does not multiply by the finite-variable value $p_0=N$.

We first prove the one-strip multiplication formula using alternants. Choose $N$ sufficiently large and let ${\beta }_i={\mu }_i+N-i$ be the decreasing beta numbers of a partition. Expanding the determinant by permutations showsIndeed, in each determinant monomial every variable occurs in exactly one row factor; multiplying that variable by its $r$th power is equivalent to raising that row exponent.

A term vanishes if the new exponent equals another exponent. Otherwise suppose the changed exponent moves from position $i$ to position $j\le i$ when restored to decreasing order. Its determinant sign is $(-1{)}^{i-j}$. The resulting partition $\lambda$ hasand all other rows unchanged. The new diagram contains $\mu$, and its size increases by $r$. Between successive added rows, the overlap is exactly the single column ${\mu }_{t-1}+1$. Thus the added cells form a connected strip without a $2\times 2$ square, of height $i-j$. Conversely every added border strip has consecutive occupied rows and exactly this single-column overlap, forcing these formulas and the corresponding exponent shift. The correspondence is therefore bijective. Divide by the unchanged Vandermonde determinant and use the bialternant formula to get the power-sum border-strip multiplication identity

Apply this formula successively to each positive part of $\alpha$. Each sequence of choices is exactly a border-strip tableau, and its product of signs is $(-1{)}^{\mathrm{ht}(T)}$. ThereforeCommutativity of the power sums also shows that the resulting coefficient does not depend on the order of the positive parts of the type.

Take $\mu$ empty and restrict to $N\ge \ell (\lambda )$ variables. Multiply by $a_{\delta }$, with $\delta =(N-1,\dots ,0)$:In an alternant $a_{\nu +\delta }$, the exponents are distinct and their decreasing arrangement is $\nu +\delta$. Consequently the monomial $x^{\lambda +\delta }$ occurs only for $\nu =\lambda$, and in that determinant its coefficient is one. This proves the Frobenius alternant character formula

To identify these coefficients with symmetric-group characters, recall the Frobenius characteristic mapThe representation-theoretic identification $\mathrm{ch}({\chi }^{\lambda })=s_{\lambda }$ can be justified by Young's rule: the permutation character on ordered blocks of sizes $\nu$ is ${\sum }_{\lambda }{K}_{\lambda \nu }{\chi }^{\lambda }$. Its characteristic is $h_{\nu }$, because its fixed block decompositions allocate whole cycles to blocks, and the exponential generating series for those allocations is ${\prod }_{j}H(t_j)$. On the symmetric-function side, Hall duality and Schur orthonormality give $h_{\nu }={\sum }_{\lambda }{K}_{\lambda \nu }{s}_{\lambda }$. Inverting the unitriangular Kostka matrix establishes the identification. Thus the coefficient of $s_{\lambda }$ in $p_{\rho }$ is the irreducible character value on cycle type $\rho$. We have supplied the border-strip computation rather than assuming the character rule.

Grouping border-strip chains by their first removal of a cycle length $r$ now yields the Murnaghan–Nakayama ruleOne may choose the $r$-cycle to be last in a strip-building chain because the power sums commute. The empty-shape value on the empty type is one, providing the recursion's initial condition.

For the staircase $\delta =(m-1,m-2,\dots ,1)$, use $m$ beta numbers including its padded zero row. Its beta set isRemoving a rim hook of length $r$ lowers one beta number by $r$ into an unoccupied nonnegative position, reversing the exponent-shift description above. For a positive even $r$, this move either hits another occupied even position or goes below zero. Therefore there are no removable even-size border strips.

Under the Hall inner product of symmetric functions, the adjoint of multiplication by $p_r$ is $r\partial /\partial p_r$: this follows on the power-sum basis from $z_{\nu }/z_{\nu \setminus r}=r{m}_r(\nu )$. Taking adjoints of the one-strip multiplication rule expresses this derivative on $s_{\delta }$ as the signed sum over removable size-$r$ strips. That sum is empty for even $r$, soThe rational symmetric-function algebra is freely generated by the power sums, and its characteristic is zero. Vanishing of each of these derivatives therefore proves the staircase Schur functions use only odd power sums statement:

Finally, the cycle type of an $n$-cycle has one strip of size $n$. A straight diagram is a border strip exactly when its second row has length at most one: otherwise it contains the upper-left $2\times 2$ square; conversely the first row with a single-column tail is connected and contains none. These diagrams are exactly the hook partitions $(n-s,1^s)$, of height $s$. There is one strip filling in this case and none otherwise, giving

A rim hook is a connected skew diagram containing no $2\times 2$ square, whose removal leaves a partition of an integer diagram. Its leg length is its number of occupied rows minus one. The Murnaghan–Nakayama rule states that, if a permutation has cycle type $(k,\rho )$, thenIteration removes one hook for each cycle length; the empty diagram has character one at the empty cycle type.

For shape $(3,3,3)$ and cycle type $(4,3,2)$, there are exactly two removable four-cell rim hooks. The remaining shapes are $(3,2)$ and $(2,2,1)$; the removed hooks have leg lengths one and two respectively. In shape $(3,2)$ the only removable three-cell rim hook leaves $(1,1)$ and has leg length one. In shape $(2,2,1)$ the only such hook leaves $(2)$ and again has leg length one. Since the characters of $(1,1)$ and $(2)$ at a transposition are $-1$ and $1$, we obtainand henceThe cycle lengths can be taken in any order; using $4,3,2$ makes the two branches of the computation explicit.

Now consider the ordinary induction product $[\alpha ][\beta ]$. A hook partition is $(x,1^y)$, equivalently a partition of an integer whose second row has length at most one. If a hook shape $\nu$ occurs in the Littlewood–Richardson rule, then $\alpha \subseteq \nu$, forcing $\alpha$ itself to be a hook. In the skew hook $\nu /\alpha$, the new cells consist of a segment in the first row and a segment down the first column, with no shared column between those two segments. Every first-row entry must be $1$: its rightmost entry is the first letter of the lattice word, hence is $1$, and weak row increase forces all entries to its left to be $1$ too. The column segment has strictly increasing entries. Thus only $1$ can be repeated, forcing the content $\beta$ to be a hook as well. This also proves the assertion when the top segment is empty, since then every entry lies in a strictly increasing column.

Write the two nonempty hook factors as $\alpha =(a,1^k)$ and $\beta =(b,1^l)$, with $k=n-r-a$ and $l=r-b$. Let the prospective resulting hook have top-row length $x$ and leg length $y$. Its new top segment has $H=x-a$ cells and its new column segment has $V=y-k$ cells. The content consists of $b$ ones and the single letters $2,\dots ,l+1$. All those larger letters must be in the new column; its strict increase allows either no $1$ or exactly one $1$. There are therefore only two possibilities:In the first case fill the top segment with $b$ ones and the column with $2,3,\dots ,l+1$. In the second, use $b-1$ top ones and column $1,2,\dots ,l+1$. Each filling is unique and has a lattice word, including the case $b=1$ with an empty top segment in the second filling. This provesBoth displayed multiplicities are one.

To deduce the full-cycle value from this product result, rather than quote the Murnaghan–Nakayama rule again, form the virtual characterFor every proper Young subgroup $S_m\times S_{n-m}$ and every pair of its irreducible characters, the hook-product calculation givesIf a factor is not a hook there are no hook constituents; if both are hooks, their two consecutive leg lengths cancel in this alternating inner product. By Frobenius reciprocity and the completeness of the product irreducible characters, $F_n$ restricts to zero on every such subgroup. Any permutation with more than one cycle preserves a nonempty proper subset and is conjugate into one of these subgroups. Thus $F_n$ is supported only on the class of $n$-cycles.

Its inner product with the trivial character is one, since only the one-row hook contributes. That class has size $(n-1)!$, so $F_n(\rho )/n=1$, giving $F_n(\rho )=n$. For any irreducible character, the support calculation then yieldsBy character orthogonality, the left side is $(-1{)}^{n-x}$ for $\nu =(x,1^{n-x})$ and zero otherwise. These are real integers, provingFor $n=1$ the same conclusion holds directly. This derives the requested special case from the Littlewood–Richardson rule and Frobenius reciprocity.

There are $(k-1)!$ $k$-cycles in $S_k$, and each has the same character value of a representation on $S^{\lambda }$. Taking traces in the cycle-sum identity for Young–Jucys–Murphy elements yieldsOnly tableaux whose first $k$ entries form a hook partition contribute.

Write such a shape as ${\mu }_b=(k-b,1^b)$, for $0\le b\le k-1$. There are $f_{{\mu }_b}=\left(\genfrac{}{}{0ex}{}{k-1}{b}\right)$ ways to fill it: choose which $b$ of the labels $2,\dots ,k$ go below the first box. There are $f^{\lambda /{\mu }_b}$ ways to fill the remaining skew diagram with the labels $k+1,\dots ,n$, in standard order. The eigenvalue from part (ii) is $(-1{)}^b(k-b-1)!b!$, so the factoris $1$. Therefore the general character value isHere $f^{\lambda /\mu }$ counts standard skew Young tableaux, and is zero when $\mu$ is not contained in $\lambda$. In particular, for a full $n$-cycle,For $k=1$, the general formula gives $f^{\lambda /(1)}=f_{\lambda }$, the character value at the identity, as it should.

Let $T$ be a standard Young tableau of shape $\lambda ⊢n$, and let $\mu$ be the shape formed by its entries $1,\dots ,k$. Since Young–Jucys–Murphy elements act by contents, the eigenvalue of ${\xi }_k$ on its Gelfand–Tsetlin basis vector isIf $\mu$ is not a hook partition, it contains $(2,2)$, whose content is zero, so this product is zero. If $\mu =(a+1,1^b)$ with $a+b+1=k$, its nonzero contents are $1,\dots ,a$ and $-1,\dots ,-b$. Their product is independent of the order of the labels. Therefore the eigenvalue isThe empty product at $k=1$ gives $1$. When $k=n$, $\mu =\lambda$, giving exactly the hook-versus-nonhook distinction suggested by the hint. For $k<n$, it is the intermediate shape $\mu$, rather than the final shape $\lambda$, that controls the answer.

A hook partition $\lambda =(n-k,1^k)$ has cell contentsIn a standard Young tableau, the unique zero-content cell is $(1,1)$ and contains $1$. The other eigenvalues are precisely the remaining contents, in whatever order the tableau supplies. Their product is independent of that order:The scalar is nonzero over the complex numbers. For the single row or single column, one of the factorials is $0!=1$; the formula still gives $(n-1)!$ or $(-1{)}^{n-1}(n-1)!$, respectively.

The Gelfand–Tsetlin basis spans $V^{\lambda }$, and the scalar computed on its vectors depends only on the shape. Thusfor every $w\in V^{\lambda }$. The product is the sum of the $(n-1)!$ permutations in the conjugacy class of an $n$-cycle. Taking traces therefore gives $(n-1)!{\chi }^{\lambda }((n))$ equal to the displayed scalar times $\dim V^{\lambda }$.

For a hook partition, a standard Young tableau is uniquely determined by the choice of its $k$ entries below the top cell, selected from $\{2,\dots ,n\}$. The column and the remaining row are then forced to increase. Hence $\dim V^{(n-k,1^k)}=\left(\genfrac{}{}{0ex}{}{n-1}{k}\right)$, and cancellation of the factorials yieldsThis uses the central character value of a conjugacy-class sum and tableau counting, without a character rule for removing rim hooks.