= Solution
The <Young permutation module> $M^{(n-k,k)}$ is the <symmetric-group subset permutation representation> on the $k$-element subsets of $\{1,\ldots,n\}$: the second row of a <tabloid> specifies the subset. For $k\geq1$, <restriction of a character> to $S_{n-1}$ fixing $n$ splits these subsets into two <orbits of a group action>, according as they contain $n$.
The corresponding <permutation representations> are on the $k$-subsets and the $(k-1)$-subsets of $\{1,\ldots,n-1\}$. Their <Young permutation modules> have two row sizes $n-1-k,k$ and $n-k,k-1$, respectively. Sort each pair into decreasing order and omit any zero part; swapping the two row labels gives an isomorphic <permutation representation>. This sorting matters at $n=2k$.
We use <Young's rule>: over the <complex numbers>, the multiplicity of $S^\rho$ in $M^\eta$ is the <Kostka number> $K_{\rho\eta}$, and it can be nonzero only if $\rho$ dominates $\eta$ in the <dominance order on partitions>. Since each $\eta$ above has at most two parts and size $n-1$, dominance gives
$$
\rho_1+\rho_2\geq\eta_1+\eta_2=n-1.
$$
Thus $\rho$ has at most two nonzero parts. The <character inner product> measures this constituent multiplicity, so
$$
\boxed{\left\langle\operatorname{Res}_{S_{n-1}}^{S_n}\pi^{(n-k,k)},\chi^\rho\right\rangle=0\quad\text{when }\ell(\rho)\geq3.}
$$
If $k=0$, the original and restricted modules are the <trivial representations>, giving the same conclusion. For $n=1$, there is no partition of $n-1=0$ with three parts, so the assertion is vacuous.
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