# Why is $h_n(1,x,\ldots,x^{m}) = h_m(1,x,\ldots,x^{n})$?

Let $$h_n$$ denote the degree $$n$$ symmetric polynomial corresponding to the homogeneous basis. That is, in, e.g., $$m$$ variables

$$h_n(x_1,\ldots,x_m) = \sum_{1\le i_1\le \cdots \le i_n \le m} \prod_{j\le n} x_{i_j}$$

Why is the following identity in $$\mathbb{Q}[x]$$ true?

$$h_n(1,x,\ldots,x^m) = h_m(1,x,\ldots,x^n)$$

I am interested in this because its equivalent to Hermite reciprocity of representations of $$\mathfrak{sl}_2$$: $$\text{Sym}^n(\text{Sym}^m(V)) \cong \text{Sym}^m(\text{Sym}^n(V))$$ for $$V$$ the standard representation.

For a partition $$\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$$ with $$\lambda_1\geq \lambda_2\geq\cdots\geq\lambda_k$$, write $$x^\lambda=x_{\lambda_1}x_{\lambda_2}\cdots x_{\lambda_k}$$. Then, with this notation, $$h_n(x_1,\ldots,x_m)=\sum_{\lambda\in X(n,m)} x^\lambda$$ where $$X(n,m)$$ is the set of partitions with $$n$$ parts of size $$\leq m$$, and $$h_n(1,x_1,\ldots,x_m)=\sum_{k=0}^{n}h_k(x_1,\ldots,x_m)=\sum_{\lambda\in\bigcup_{k=0}^nX(k,m)}x^\lambda$$ Similarly, $$h_m(1,x_1,\ldots,x_n)=\sum_{\lambda\in \bigcup_{k=0}^m X(k,n)} x^\lambda.$$ Note that the transpose map $$\lambda\to\lambda^t$$ defines a bijection $$\bigcup_{k=0}^mX(k,n)\to \bigcup_{k=0}^nX(k,m)$$.
Now, for $$\lambda=(\lambda_1,\lambda_2,\ldots,\lambda_k)$$, set $$|\lambda|=\lambda_1+\cdots+\lambda_k$$ and note that $$|\lambda|=|\lambda^t|$$. Therefore, \begin{align} h_n(1,x,x^2,\ldots,x^m)&=\sum_{\lambda\in \bigcup_{k=0}^nX(k,m)} x^{|\lambda|}\\ &=\sum_{\lambda\in \bigcup_{k=0}^mX({k,n})}x^{|\lambda^t|}\\ &=\sum_{\lambda\in \bigcup_{k=0}^mX({k,n})}x^{|\lambda|}\\ &=h_m(1,x,x^2,\ldots,x^{n}). \end{align}