[proofplan]
We prove the identity by applying the standard involution $\omega$ of the symmetric function ring to the ordinary Jacobi-Trudi formula for the conjugate partition $\lambda'$. The ordinary Jacobi-Trudi formula expresses $s_{\lambda'}$ as a determinant in the complete homogeneous symmetric functions $h_r$. The involution $\omega$ sends $h_r$ to $e_r$ and sends $s_{\lambda'}$ to $s_\lambda$, so applying $\omega$ entrywise to that determinant gives exactly the displayed determinant.
[/proofplan]
[step:Apply the ordinary Jacobi-Trudi identity to the conjugate partition]
Let $\mu := \lambda'$ be the conjugate partition of $\lambda$. Since $m \geq \lambda_1$ and the length of $\lambda'$ is $\lambda_1$, the integer $m$ is at least the number of nonzero parts of $\mu$. Therefore the ordinary [Jacobi-Trudi identity](/theorems/5181) applies to $\mu$ with this value of $m$ and gives
\begin{align*}
s_{\lambda'} = s_\mu
= \det\bigl(h_{\mu_i - i + j}\bigr)_{1 \leq i,j \leq m}
= \det\bigl(h_{\lambda'_i - i + j}\bigr)_{1 \leq i,j \leq m},
\end{align*}
where $h_0 = 1$ and $h_r = 0$ for $r < 0$.
Here we are citing a result not yet in the wiki: ordinary Jacobi-Trudi identity.
[guided]
We want the determinant in the theorem to involve the parts of $\lambda'$, so the correct input for the ordinary Jacobi-Trudi identity is the partition $\mu := \lambda'$. The ordinary Jacobi-Trudi identity says that if $m$ is at least the length of a partition $\mu$, then
\begin{align*}
s_\mu = \det\bigl(h_{\mu_i - i + j}\bigr)_{1 \leq i,j \leq m}.
\end{align*}
We must verify the size condition. The length of the conjugate partition $\lambda'$ is exactly $\lambda_1$, because the first row of $\lambda$ has $\lambda_1$ boxes, and these boxes correspond to the nonzero columns of $\lambda$. Since the theorem assumes $m \geq \lambda_1$, we have $m \geq \ell(\lambda') = \ell(\mu)$. Hence the Jacobi-Trudi identity applies with $\mu = \lambda'$, giving
\begin{align*}
s_{\lambda'} = s_\mu
= \det\bigl(h_{\mu_i - i + j}\bigr)_{1 \leq i,j \leq m}
= \det\bigl(h_{\lambda'_i - i + j}\bigr)_{1 \leq i,j \leq m}.
\end{align*}
This is the point at which the hypothesis $m \geq \lambda_1$ is used: it ensures that the determinant has enough rows and columns for the Jacobi-Trudi formula of the conjugate partition.
Here we are citing a result not yet in the wiki: ordinary Jacobi-Trudi identity.
[/guided]
[/step]
[step:Apply the omega involution to the determinant]
Let
\begin{align*}
\omega: \operatorname{Sym} &\to \operatorname{Sym}
\end{align*}
be the standard involutive ring automorphism determined by
\begin{align*}
\omega(h_r) = e_r
\end{align*}
for every integer $r \geq 0$. Since $\omega$ is a ring homomorphism and the determinant is a finite signed sum of products, applying $\omega$ to the Jacobi-Trudi determinant gives
\begin{align*}
\omega(s_{\lambda'})
&= \omega\left(\det\bigl(h_{\lambda'_i - i + j}\bigr)_{1 \leq i,j \leq m}\right) \\
&= \det\bigl(\omega(h_{\lambda'_i - i + j})\bigr)_{1 \leq i,j \leq m} \\
&= \det\bigl(e_{\lambda'_i - i + j}\bigr)_{1 \leq i,j \leq m}.
\end{align*}
The conventions are preserved because $\omega(h_0) = e_0 = 1$ and, for negative indices, both $h_r$ and $e_r$ are defined to be $0$.
[/step]
[step:Identify the omega image of the conjugate Schur function]
The standard conjugation property of Schur functions under $\omega$ states that
\begin{align*}
\omega(s_\nu) = s_{\nu'}
\end{align*}
for every partition $\nu$. Applying this with $\nu = \lambda'$ gives
\begin{align*}
\omega(s_{\lambda'}) = s_{(\lambda')'} = s_\lambda.
\end{align*}
Here we are citing a result not yet in the wiki: the omega involution sends Schur functions to Schur functions of conjugate shape.
[/step]
[step:Combine the two identities]
From the determinant identity obtained after applying $\omega$ and the conjugation property of Schur functions, we have
\begin{align*}
s_\lambda
= \omega(s_{\lambda'})
= \det\bigl(e_{\lambda'_i - i + j}\bigr)_{1 \leq i,j \leq m}.
\end{align*}
This is exactly the desired dual Jacobi-Trudi identity.
[/step]
