[proofplan]
We first fix the standard embedding $S_{n-1}\le S_n$ and use the [Young branching rule for restriction](/theorems/8443), which identifies the restriction of a Specht module with one summand for each removable box. Since the field has characteristic zero, Maschke semisimplicity upgrades the branching filtration to a direct-sum decomposition. The induction formula is then the corresponding [Young branching rule for induction](/theorems/8444), equivalently the adjoint statement obtained from the same multiplicity-free branching pattern.
[/proofplan]
[step:Apply the restriction branching rule to the standard subgroup fixing $n$]
Let $S_{n-1}\le S_n$ denote the subgroup fixing $n$. For a partition $\nu\vdash n$, let $\mathcal R(\nu)$ denote the set of removable boxes of the Young diagram of $\nu$. For each box $b\in \mathcal R(\nu)$, define $\nu\setminus b$ to be the partition of $n-1$ obtained by deleting $b$.
By [Young Branching Rule For Restriction]([citetheorem:8443]), applied over the characteristic-zero field $k$ and to the standard subgroup $S_{n-1}\le S_n$ fixing $n$, the restricted module $\operatorname{Res}^{S_n}_{S_{n-1}}S^\nu$ has a Specht filtration whose successive quotients are precisely the modules
\begin{align*}
S^{\nu\setminus b}
\end{align*}
as $b$ ranges over $\mathcal R(\nu)$.
[guided]
The subgroup matters: throughout the argument, $S_{n-1}$ is the copy of the symmetric group inside $S_n$ that fixes the entry $n$ and permutes $\{1,\dots,n-1\}$. This is exactly the subgroup used in the tableau model of Specht modules, because one can track where the entry $n$ occurs in a standard tableau.
Let $\mathcal R(\nu)$ be the set of removable boxes of the Young diagram of $\nu$. For $b\in\mathcal R(\nu)$, write $\nu\setminus b$ for the partition of $n-1$ obtained by deleting that box. The restriction branching theorem [Young Branching Rule For Restriction]([citetheorem:8443]) applies because the field $k$ has characteristic zero, $\nu$ is a partition of $n$, and the subgroup under consideration is precisely the standard subgroup $S_{n-1}\le S_n$ fixing $n$. It gives a filtration of the restricted $kS_{n-1}$-module
\begin{align*}
\operatorname{Res}^{S_n}_{S_{n-1}}S^\nu
\end{align*}
whose successive quotients are the Specht modules
\begin{align*}
S^{\nu\setminus b}
\end{align*}
for the partitions obtained by removing one removable box from $\nu$.
The tableau reason behind this filtration is that, in the standard-polytabloid model, the position of the entry $n$ determines a removable box. Deleting that box leaves a standard tableau of the smaller shape, and the remaining entries are acted on by $S_{n-1}$.
[/guided]
[/step]
[step:Split the restriction filtration using semisimplicity]
Since $\operatorname{char}(k)=0$, the characteristic of $k$ does not divide $|S_{n-1}|$. By [Maschke Theorem For Finite Groups]([citetheorem:8439]), every finite-dimensional $kS_{n-1}$-module is completely reducible. Therefore the Specht filtration from the previous step splits as a direct sum, and we obtain an isomorphism of $kS_{n-1}$-modules
\begin{align*}
\operatorname{Res}^{S_n}_{S_{n-1}} S^\nu \cong \bigoplus_{b\in\mathcal R(\nu)} S^{\nu\setminus b}.
\end{align*}
Reindexing the summands by partitions $\mu\vdash n-1$ satisfying $\mu\nearrow\nu$ gives
\begin{align*}
\operatorname{Res}^{S_n}_{S_{n-1}} S^\nu \cong \bigoplus_{\mu\nearrow\nu} S^\mu.
\end{align*}
[/step]
[step:Apply the induction branching rule for addable boxes]
Now let $\mu\vdash n-1$. By [Young Branching Rule for Induction]([citetheorem:8444]), applied over the same characteristic-zero field $k$ and to the standard inclusion $S_{n-1}\le S_n$, the induced module decomposes as
\begin{align*}
\operatorname{Ind}^{S_n}_{S_{n-1}}S^\mu \cong \bigoplus_{\mu\nearrow\nu} S^\nu,
\end{align*}
where $\nu$ ranges over the partitions of $n$ obtained from $\mu$ by adding one addable box. This is exactly the asserted induction formula.
[/step]
[step:Identify the two indexing descriptions]
The condition that $\mu\nearrow\nu$ means that the Young diagram of $\nu$ is obtained from that of $\mu$ by adding one box. Equivalently, $\mu$ is obtained from $\nu$ by removing the same box, and that box is removable in $\nu$. Thus the restriction formula is indexed by removable boxes of $\nu$, while the induction formula is indexed by addable boxes of $\mu$. These are precisely the two decompositions stated in the theorem.
[/step]
