[proofplan]
We prove the identity by testing both sides against every element of the power-sum basis. The adjoint definition converts the left-hand pairing into the Hall [inner product](/page/Inner%20Product) of $p_\lambda$ with $p_kp_\mu$, and power-sum orthogonality makes this nonzero exactly when $\lambda$ is obtained from $\mu$ by adjoining one part $k$. In that case the scalar is computed from the explicit formula for $z_\lambda$, and nondegeneracy of the Hall inner product on the power-sum basis identifies the two symmetric functions.
[/proofplan]
[step:Pair the adjoint against an arbitrary power-sum basis element]
Let $\mu=(1^{m_1(\mu)}2^{m_2(\mu)}\cdots)$ be an arbitrary partition. Since $p_k^\perp$ is the adjoint of the multiplication map
\begin{align*}
M_{p_k}:\Lambda_{\mathbb{Q}} &\to \Lambda_{\mathbb{Q}}\\
f &\mapsto p_k f,
\end{align*}
we have
\begin{align*}
\langle p_k^\perp p_\lambda, p_\mu\rangle
=
\langle p_\lambda, p_k p_\mu\rangle.
\end{align*}
The product of power sums is indexed by multiset union of parts, so
\begin{align*}
p_k p_\mu = p_{\mu\cup k},
\end{align*}
where $\mu\cup k$ denotes the partition obtained from $\mu$ by adjoining one part equal to $k$. Therefore the defining orthogonality relation for the Hall inner product gives
\begin{align*}
\langle p_k^\perp p_\lambda, p_\mu\rangle
=
\langle p_\lambda, p_{\mu\cup k}\rangle
=
\begin{cases}
z_\lambda, & \lambda=\mu\cup k,\\
0, & \lambda\neq \mu\cup k.
\end{cases}
\end{align*}
[guided]
We test against $p_\mu$ because the family $\{p_\mu\}$ is the basis in which the Hall inner product is diagonal. Let $\mu=(1^{m_1(\mu)}2^{m_2(\mu)}\cdots)$ be an arbitrary partition. The operator $p_k^\perp$ is defined as the adjoint of multiplication by $p_k$, meaning precisely that for every $f,g\in\Lambda_{\mathbb{Q}}$,
\begin{align*}
\langle p_k^\perp f,g\rangle=\langle f,p_k g\rangle.
\end{align*}
Applying this with $f=p_\lambda$ and $g=p_\mu$ gives
\begin{align*}
\langle p_k^\perp p_\lambda, p_\mu\rangle
=
\langle p_\lambda, p_k p_\mu\rangle.
\end{align*}
Now multiplication of power sums simply concatenates parts. If $\mu$ has multiplicities $m_i(\mu)$, then $p_kp_\mu$ has multiplicities $m_k(\mu)+1$ at $k$ and $m_i(\mu)$ at every $i\neq k$. Thus
\begin{align*}
p_kp_\mu=p_{\mu\cup k},
\end{align*}
where $\mu\cup k$ is the partition obtained by adjoining one part $k$.
The Hall inner product is diagonal on the power-sum basis:
\begin{align*}
\langle p_\alpha,p_\beta\rangle=\delta_{\alpha,\beta}z_\alpha.
\end{align*}
Using this with $\alpha=\lambda$ and $\beta=\mu\cup k$ yields
\begin{align*}
\langle p_k^\perp p_\lambda, p_\mu\rangle
=
\langle p_\lambda, p_{\mu\cup k}\rangle
=
\begin{cases}
z_\lambda, & \lambda=\mu\cup k,\\
0, & \lambda\neq \mu\cup k.
\end{cases}
\end{align*}
This is the whole point of passing to pairings: the adjoint moves $p_k^\perp$ off the unknown expression, and orthogonality reduces the computation to a single possible partition.
[/guided]
[/step]
[step:Compute the pairing of the proposed right-hand side]
First suppose $m_k(\lambda)=0$. Then no partition $\mu$ satisfies $\lambda=\mu\cup k$, so the previous step gives
\begin{align*}
\langle p_k^\perp p_\lambda,p_\mu\rangle=0
\end{align*}
for every partition $\mu$, matching the proposed value $0$.
Now suppose $m_k(\lambda)>0$, and let $\lambda\setminus k$ be the partition whose multiplicities are
\begin{align*}
m_i(\lambda\setminus k)
=
\begin{cases}
m_k(\lambda)-1, & i=k,\\
m_i(\lambda), & i\neq k.
\end{cases}
\end{align*}
The proposed right-hand side is
\begin{align*}
R:=k\,m_k(\lambda)\,p_{\lambda\setminus k}.
\end{align*}
For every partition $\mu$,
\begin{align*}
\langle R,p_\mu\rangle
=
k\,m_k(\lambda)\,\langle p_{\lambda\setminus k},p_\mu\rangle
=
\begin{cases}
k\,m_k(\lambda)\,z_{\lambda\setminus k}, & \mu=\lambda\setminus k,\\
0, & \mu\neq \lambda\setminus k.
\end{cases}
\end{align*}
Using the definition of $z_\lambda$ and separating the factor indexed by $k$,
\begin{align*}
z_\lambda
&=
\prod_{i\geq 1} i^{m_i(\lambda)}m_i(\lambda)!\\
&=
k^{m_k(\lambda)}m_k(\lambda)!\prod_{i\neq k}i^{m_i(\lambda)}m_i(\lambda)!\\
&=
k\,m_k(\lambda)\,
k^{m_k(\lambda)-1}(m_k(\lambda)-1)!\prod_{i\neq k}i^{m_i(\lambda)}m_i(\lambda)!\\
&=
k\,m_k(\lambda)\,z_{\lambda\setminus k}.
\end{align*}
Thus
\begin{align*}
\langle R,p_\mu\rangle
=
\begin{cases}
z_\lambda, & \mu=\lambda\setminus k,\\
0, & \mu\neq \lambda\setminus k.
\end{cases}
\end{align*}
Since $\lambda=\mu\cup k$ is equivalent to $\mu=\lambda\setminus k$ when $m_k(\lambda)>0$, this is exactly the same pairing computed for $p_k^\perp p_\lambda$.
[/step]
[step:Use nondegeneracy of the Hall inner product to identify the symmetric functions]
If $m_k(\lambda)=0$, the first step gives
\begin{align*}
\langle p_k^\perp p_\lambda,p_\mu\rangle=0
\end{align*}
for every partition $\mu$. Since the power sums form a basis of $\Lambda_{\mathbb{Q}}$ and the Hall inner product satisfies $\langle p_\mu,p_\mu\rangle=z_\mu\neq 0$ for every partition $\mu$, the inner product is nondegenerate. Hence $p_k^\perp p_\lambda=0$.
If $m_k(\lambda)>0$, the previous step shows that
\begin{align*}
\langle p_k^\perp p_\lambda,p_\mu\rangle
=
\langle k\,m_k(\lambda)\,p_{\lambda\setminus k},p_\mu\rangle
\end{align*}
for every partition $\mu$. By the same nondegeneracy, the two symmetric functions are equal:
\begin{align*}
p_k^\perp p_\lambda
=
k\,m_k(\lambda)\,p_{\lambda\setminus k}.
\end{align*}
Combining the two cases proves the stated formula.
[/step]
