= Solution
For a <Young diagram> $\lambda$, a <Removable node of a Young diagram> is a cell whose deletion leaves a <partition of an integer> diagram; equivalently it is the last cell of row $i$ with $\lambda_i>\lambda_{i+1}$. Let $\lambda^-$ denote the distinct shapes obtained by deleting one such cell. The <restriction branching rule for a symmetric group> in <characteristic zero> is
$$
\boxed{S^\lambda\downarrow_{S_{n-1}}
\cong\bigoplus_{\mu\in\lambda^-}S^\mu.}
$$
Its induction version adds one cell in each possible way. Over an arbitrary <field>, restriction instead has a <Specht filtration> with these successive quotients, each once; it need not split.
To prove restriction, list the removable nodes from top to bottom. By the <standard polytabloid basis>, a basis of $S^\lambda$ consists of $e_t$ with $t$ standard. Its entry $n$ is necessarily at a removable node. Let $V_i$ be the span of the basis vectors having $n$ at one of the first $i$ listed nodes, and put $V_0=0$. These are $S_{n-1}$-invariant. To verify this, apply a <permutation> of $1,\ldots,n-1$ and straighten the resulting <polytabloid> by column antisymmetry and <Garnir relations>. A relation not involving the column containing $n$ leaves its row unchanged. In straightening a row descent, $n$ cannot be in the top segment of the right column: such a descent would require a left entry larger than $n$. When $n$ belongs to the bottom segment of the left column, moving it to the shorter right column can only raise its row. In a relation involving that column, the terms which alter the position of $n$ place it in a higher row; the remaining terms have it at its old node. Thus straightening never moves it past the current top-to-bottom cutoff.
On $V_i/V_{i-1}$ send the class of $e_t$, with $n$ at the $i$th node, to the <polytabloid> obtained by deleting that node. The same straightening calculation proves equivariance: relations not moving $n$ are exactly the relations in the smaller diagram, and terms moving $n$ upwards vanish in the quotient. Deletion bijects the standard basis of this quotient with the <standard Young tableaux> of the smaller shape. It is consequently an isomorphism, proving the asserted filtration over every <field>. In <characteristic zero>, <Maschke's theorem> splits the filtration and gives the direct sum. Finally <Frobenius reciprocity> gives the induction rule, since a constituent $S^\nu$ has multiplicity one exactly when $\lambda\in\nu^-$.
The <hook-length formula> states
$$
\boxed{f^\lambda=\dim S^\lambda=\frac{n!}{\prod_{(i,j)\in\lambda}h_{ij}},
\qquad h_{ij}=\lambda_i-j+\lambda'_j-i+1.}
$$
The <hook of a Young diagram> consists of its cell, the cells to its right in the same row, and the cells below it in the same column. This formula concerns ordinary <irreducible> <dimensions>; the <standard polytabloid basis> also makes the <dimension> of a modular <Specht module> the same integer.
All <transpositions> are conjugate, have order two, and generate $S_n$. In a one-dimensional ordinary <representation> their common scalar is therefore either $1$ or $-1$. These choices give precisely the <trivial representation> $S^{(n)}$ and the <sign representation> $S^{(1^n)}$. At $n=1$ they coincide. The standard shape $(n-1,1)$ has <dimension> $n-1$: its <standard Young tableau> is specified by choosing the bottom entry from $2,\ldots,n$. Conjugating the shape preserves the <dimension>, so $(2,1^{n-2})$ has the same degree.
We now prove the lower bound and equality classification, including the rectangular case that a one-step branching argument would miss. Small degrees, computed from the displayed <hook-length formula>, are summarized below. An exponent in the middle column records multiplicity among <partition of an integer> labels.
|| $n$
|| Complete multiset of <irreducible> degrees
|| Least degree of a shape other than the linear and standard shapes
| $3$
| $1^{(2)},2$
| No other shape
| $4$
| $1^{(2)},2,3^{(2)}$
| $2$ at $(2,2)$
| $5$
| $1^{(2)},4^{(2)},5^{(2)},6$
| $5$
| $6$
| $1^{(2)},5^{(4)},9^{(2)},10^{(2)},16$
| $5$ at $(3,3),(2,2,2)$
| $7$
| $1^{(2)},6^{(2)},14^{(4)},15^{(2)},20,21^{(2)},35^{(2)}$
| $14$
| $8$
| $1^{(2)},7^{(2)},14^{(2)},20^{(2)},21^{(2)},28^{(2)},35^{(2)},42,56^{(2)},64^{(2)},70^{(2)},90$
| $14$
For example, at $n=6$ the conjugate pairs $(6),(1^6)$, $(5,1),(2,1^4)$, $(4,2),(2,2,1,1)$, $(4,1,1),(3,1,1,1)$, $(3,3),(2,2,2)$ have degrees $1,5,9,10,5$, and the remaining shape $(3,2,1)$ has degree $16$. These checks also identify every small equality case.
Induct on $n\ge9$, assuming the lower bound for all smaller degrees. If a non-linear, nonstandard shape has at least two removable nodes, none of its predecessors is a single row or single column: adding one cell to a row or column could produce only a linear or standard shape. Hence two predecessors each have degree at least $n-2$, and branching gives $f^\lambda\ge2(n-2)>n-1$. If it has only one removable node, it is a rectangle $(a^b)$ with $a,b\ge2$. Its sole predecessor is $(a^{b-1},a-1)$, which has two removable nodes. Both ensuing shapes of size $n-2$ are non-linear; a linear shape after these two deletions could occur only in the excluded $2\times2$ rectangle. Therefore two-step branching gives $f^\lambda\ge2(n-3)>n-1$. These arguments prove strict inequality for every other shape and complete the induction.
Consequently the <least non-linear ordinary degree of a symmetric group> is
$$
\boxed{n-1\text{ for }n\ge3,\ n\ne4;qquad 2\text{ for }n=4.}
$$
For degree $n-1$, the only labels are the two standard conjugate shapes, except that at $n=6$ there are also $(3,3)$ and $(2,2,2)$, both degree five. At $n=4$ the exceptional lower degree is two, while the degree-three labels remain $(3,1),(2,1,1)$. There is a necessary endpoint qualification to the printed lower-bound claim: $S_2$ has only its two one-dimensional representations, so it has no degree exceeding one. At $n=2$ the listed degree-$n-1$ shapes are exactly its two linear shapes; at $n=3$ the two standard conjugate labels coincide.
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