# How to prove $\prod_{\lambda\vdash n}\prod_im_i(\lambda)!=\prod_{\lambda\vdash n}1^{m_1(\lambda)}2^{m_2(\lambda)}\cdots$

Let $\lambda$ be a partition of an positive integer $n$, it can be presented as $\lambda=(\lambda_{1},\lambda_2,\cdots,\lambda_l)$ such that $\lambda_1\geq\lambda_2\geq\cdots\geq\lambda_l>0$, or $\lambda=(1^{m_1},2^{m_2},\cdots,k^{m_k})$, here $m_i$ is the number of times that the part $i$ appears in the partition $\lambda$. Apparently, in this question, we use the second form.

how to prove $\prod\limits_{\lambda\vdash n}\prod\limits_im_i(\lambda)!=\prod\limits_{\lambda\vdash n}1^{m_1(\lambda)}2^{m_2(\lambda)}\cdots$? It is clear that the right side of the identity is the product of all parts of all partitions of $n$.

I have seen this problem in Enumerative Combinatorics, Vol.2 of R.P.Stanley, Chapter 7, Symmetric Functions,Section 7.6, An Involution. And any combinatorial interpretation is more expected.