[proofplan]
We use the Hall [inner product](/page/Inner%20Product) on $\Lambda$, for which the Schur functions form an [orthonormal basis](/page/Orthonormal%20Basis). The skew Schur function $s_{\nu/\lambda}$ is characterized by adjointness: taking the Hall inner product with $s_\mu$ is the same as taking the Hall inner product of $s_\nu$ with $s_\lambda s_\mu$. Expanding $s_\lambda s_\mu$ in the Schur basis then identifies this scalar product with the Littlewood-Richardson coefficient $c_{\lambda\mu}^{\nu}$. Since coefficients in an orthonormal basis are obtained by inner products with the basis vectors, these scalar products give exactly the desired expansion.
[/proofplan]
[step:Extract Schur coefficients using the Hall inner product]
Let $(\cdot,\cdot)$ denote the Hall inner product on $\Lambda$, normalized by
\begin{align*}
(s_\alpha,s_\beta)=\delta_{\alpha\beta}
\end{align*}
for all partitions $\alpha$ and $\beta$, where $\delta_{\alpha\beta}$ is the Kronecker delta. Since the Schur functions form an orthonormal basis of $\Lambda$, every homogeneous symmetric function $f \in \Lambda$ has the expansion
\begin{align*}
f=\sum_{\mu}(f,s_\mu)s_\mu,
\end{align*}
where the sum ranges over partitions $\mu$ of the degree of $f$. Applying this to $f=s_{\nu/\lambda}$, it is enough to prove that
\begin{align*}
(s_{\nu/\lambda},s_\mu)=c_{\lambda\mu}^{\nu}
\end{align*}
for every partition $\mu$.
[guided]
The purpose of the Hall inner product is to turn the problem of finding Schur coefficients into a problem about scalar products. We write $(\cdot,\cdot)$ for the Hall inner product on $\Lambda$, and its defining normalization on the Schur basis is
\begin{align*}
(s_\alpha,s_\beta)=\delta_{\alpha\beta}
\end{align*}
for all partitions $\alpha$ and $\beta$. Thus the Schur basis is orthonormal.
For any homogeneous symmetric function $f \in \Lambda$, the coefficient of $s_\mu$ in its Schur expansion is obtained by pairing $f$ with $s_\mu$. Indeed, if
\begin{align*}
f=\sum_{\rho} a_\rho s_\rho
\end{align*}
for scalars $a_\rho$, then orthonormality gives
\begin{align*}
(f,s_\mu)
=
\left(\sum_{\rho} a_\rho s_\rho,s_\mu\right)
=
\sum_{\rho} a_\rho(s_\rho,s_\mu)
=
\sum_{\rho} a_\rho\delta_{\rho\mu}
=
a_\mu.
\end{align*}
Therefore, to prove
\begin{align*}
s_{\nu/\lambda}=\sum_{\mu} c_{\lambda\mu}^{\nu}s_\mu,
\end{align*}
it is enough to show that each Schur coefficient of $s_{\nu/\lambda}$ is $c_{\lambda\mu}^{\nu}$, namely that
\begin{align*}
(s_{\nu/\lambda},s_\mu)=c_{\lambda\mu}^{\nu}
\end{align*}
for every partition $\mu$.
[/guided]
[/step]
[step:Use adjointness to move skewing onto multiplication by $s_\lambda$]
By the defining adjointness property of skew Schur functions, multiplication by $s_\lambda$ is adjoint to skewing by $\lambda$. Hence, for every partition $\mu$,
\begin{align*}
(s_{\nu/\lambda},s_\mu)=(s_\nu,s_\lambda s_\mu).
\end{align*}
[/step]
[step:Identify the resulting scalar product with the Littlewood-Richardson coefficient]
By definition of the Littlewood-Richardson coefficients, the product $s_\lambda s_\mu$ expands as
\begin{align*}
s_\lambda s_\mu=\sum_{\rho}c_{\lambda\mu}^{\rho}s_\rho,
\end{align*}
where the sum ranges over all partitions $\rho$. Pairing both sides with $s_\nu$ and using orthonormality of the Schur basis gives
\begin{align*}
(s_\nu,s_\lambda s_\mu)
&=
\left(s_\nu,\sum_{\rho}c_{\lambda\mu}^{\rho}s_\rho\right) \\
&=
\sum_{\rho}c_{\lambda\mu}^{\rho}(s_\nu,s_\rho) \\
&=
\sum_{\rho}c_{\lambda\mu}^{\rho}\delta_{\nu\rho} \\
&=
c_{\lambda\mu}^{\nu}.
\end{align*}
Combining this with the adjointness identity yields
\begin{align*}
(s_{\nu/\lambda},s_\mu)=c_{\lambda\mu}^{\nu}
\end{align*}
for every partition $\mu$.
[/step]
[step:Reconstruct the skew Schur function from its Schur coefficients]
Using the Schur coefficient extraction formula from the Hall inner product, we obtain
\begin{align*}
s_{\nu/\lambda}
=
\sum_{\mu}(s_{\nu/\lambda},s_\mu)s_\mu
=
\sum_{\mu}c_{\lambda\mu}^{\nu}s_\mu.
\end{align*}
This is the claimed Schur expansion of the skew Schur function $s_{\nu/\lambda}$.
[/step]
