[proofplan] We use the standard Specht-module branching filtration on the restriction of $S^\lambda$ to $S_{n-1}$. The filtration is indexed by the removable boxes of $\lambda$, ordered from top to bottom, and its successive quotients are Specht modules for the partitions obtained by deleting those boxes. Since $k$ has characteristic zero, [Maschke's theorem](/theorems/2409) makes $kS_{n-1}$ semisimple, so the filtration splits. This produces the asserted direct sum, with one summand for each removable box. [/proofplan] [step:Order the removable boxes and state the Specht branching filtration] Let $b_1,\dots,b_r$ be the removable boxes of the Young diagram of $\lambda$, ordered from top row to bottom row. For each $j \in \{1,\dots,r\}$, define $\mu_j := \lambda \setminus b_j$, so $\mu_j \vdash n-1$ is the partition obtained by deleting $b_j$ from $\lambda$. Thus the set of partitions $\mu$ satisfying $\mu \nearrow \lambda$ is exactly $\{\mu_1,\dots,\mu_r\}$. We use the Specht branching filtration theorem for restriction, applied to the partition $\lambda \vdash n$ and to the standard subgroup inclusion $S_{n-1}\le S_n$ fixing $n$. Its hypotheses are satisfied here because $S^\lambda$ is the Specht $kS_n$-module associated to $\lambda$, the removable boxes $b_1,\dots,b_r$ have been ordered from top row to bottom row, and the restriction is taken along that standard embedded copy of $S_{n-1}$. The theorem gives $kS_{n-1}$-submodules \begin{align*} 0 = F_0 \subset F_1 \subset \cdots \subset F_r = \operatorname{Res}^{S_n}_{S_{n-1}} S^\lambda \end{align*} such that, for every $j \in \{1,\dots,r\}$, \begin{align*} F_j/F_{j-1} \cong S^{\mu_j} \end{align*} as $kS_{n-1}$-modules. The filtration theorem is the standard Specht-module restriction result proved from the polytabloid construction: using the [Polytabloid Realisation Of Specht Modules]([citetheorem:8441]) and the [Standard Polytabloid Basis Theorem]([citetheorem:8442]), one orders standard tableaux by the removable box occupied by the entry $n$ and takes the spans of the corresponding initial segments. The theorem includes the two non-formal combinatorial assertions needed here: these initial-segment spans are stable under the subgroup $S_{n-1}$ fixing $n$, and the $j$th successive quotient is naturally identified with the Specht module of the shape obtained by deleting $b_j$. Thus no additional straightening argument is being assumed in this proof beyond the cited polytabloid construction and the branching-filtration theorem. [/step] [step:Split the filtration using semisimplicity of $kS_{n-1}$] Because $k$ has characteristic zero, the characteristic of $k$ does not divide \begin{align*} |S_{n-1}| = (n-1)!. \end{align*} By [Maschke Theorem For Finite Groups]([citetheorem:8439]), every finite-dimensional $kS_{n-1}$-module is completely reducible. In particular, every short exact sequence of finite-dimensional $kS_{n-1}$-modules splits. Apply this to the short exact sequence \begin{align*} 0 \longrightarrow F_{r-1} \longrightarrow F_r \longrightarrow F_r/F_{r-1} \longrightarrow 0. \end{align*} It follows that \begin{align*} F_r \cong F_{r-1} \oplus F_r/F_{r-1}. \end{align*} Repeating this splitting for the short exact sequences \begin{align*} 0 \longrightarrow F_{j-1} \longrightarrow F_j \longrightarrow F_j/F_{j-1} \longrightarrow 0 \end{align*} for $j=r,r-1,\dots,1$, we obtain \begin{align*} \operatorname{Res}^{S_n}_{S_{n-1}} S^\lambda = F_r \cong \bigoplus_{j=1}^r F_j/F_{j-1}. \end{align*} [guided] Recall the objects supplied by the previous step. The boxes $b_1,\dots,b_r$ are the removable boxes of the Young diagram of $\lambda$, ordered from top row to bottom row. For each $j\in\{1,\dots,r\}$, the partition $\mu_j:=\lambda\setminus b_j$ is obtained from $\lambda$ by deleting $b_j$. The Specht branching filtration theorem gives the restricted $kS_{n-1}$-module $\operatorname{Res}^{S_n}_{S_{n-1}}S^\lambda$ a filtration by $kS_{n-1}$-submodules \begin{align*} 0 = F_0 \subset F_1 \subset \cdots \subset F_r = \operatorname{Res}^{S_n}_{S_{n-1}} S^\lambda \end{align*} with quotient identifications \begin{align*} F_j/F_{j-1}\cong S^{\mu_j} \end{align*} for every $j\in\{1,\dots,r\}$. This filtration already contains the representation-theoretic information, but it gives it in successive layers rather than as a direct sum. To turn layers into summands, we need a splitting theorem. The group acting after restriction is $S_{n-1}$, whose order is \begin{align*} |S_{n-1}| = (n-1)!. \end{align*} Since $k$ has characteristic zero, no positive integer is zero in $k$, so $\operatorname{char}(k)$ does not divide $(n-1)!$. Therefore the hypotheses of [Maschke Theorem For Finite Groups]([citetheorem:8439]) are satisfied for the finite group $S_{n-1}$ and the field $k$. The conclusion is that every finite-dimensional $kS_{n-1}$-module is completely reducible. Equivalently, every submodule of such a module has a complementary submodule. Apply this first to the inclusion $F_{r-1} \subset F_r$. The quotient map $F_r \to F_r/F_{r-1}$ is a homomorphism of $kS_{n-1}$-modules, and the exact sequence \begin{align*} 0 \longrightarrow F_{r-1} \longrightarrow F_r \longrightarrow F_r/F_{r-1} \longrightarrow 0 \end{align*} therefore splits. Hence \begin{align*} F_r \cong F_{r-1} \oplus F_r/F_{r-1}. \end{align*} Now apply the same argument to $F_{r-2} \subset F_{r-1}$, then to $F_{r-3} \subset F_{r-2}$, and continue down the filtration. At the end, all filtration layers have split off, giving \begin{align*} F_r \cong \bigoplus_{j=1}^r F_j/F_{j-1}. \end{align*} Since $F_r$ is the restricted module $\operatorname{Res}^{S_n}_{S_{n-1}} S^\lambda$, this is exactly the direct-sum form of the branching rule. [/guided] [/step] [step:Identify the split summands with the Specht modules indexed by deleted boxes] From the branching filtration, each quotient satisfies \begin{align*} F_j/F_{j-1} \cong S^{\mu_j} \end{align*} as a $kS_{n-1}$-module. Substituting these quotient identifications into the split decomposition gives \begin{align*} \operatorname{Res}^{S_n}_{S_{n-1}} S^\lambda \cong \bigoplus_{j=1}^r S^{\mu_j}. \end{align*} The partitions $\mu_1,\dots,\mu_r$ are precisely the partitions $\mu \vdash n-1$ obtained from $\lambda$ by deleting one removable box. Thus \begin{align*} \operatorname{Res}^{S_n}_{S_{n-1}} S^\lambda \cong \bigoplus_{\mu \nearrow \lambda} S^\mu. \end{align*} Each removable box contributes exactly one quotient in the filtration, because in any standard tableau of shape $\lambda$ the largest entry $n$ must occupy a removable box, and it occupies exactly one box. This proves the claimed branching rule. [/step]