Denote an integer partition of $n$ by $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k)$ where $\lambda_k>0$. Also recall the $q$-analogues of integer $n$ given by $[n]_q=\frac{1-q^n}{1-q}$. Further, let $$[n]_q!=[n]_q[n-1]_q\cdots[2]_q[1]_q \qquad \text{and} \qquad [0]_q!=1.$$ If $\lambda=(\lambda_1\geq\lambda_2\geq\dots\geq\lambda_k)\vdash n$, define $$a(\lambda):=[\lambda_k]_q!\prod_{j=1}^{k-1}\,\,[\lambda_j-\lambda_{j+1}]_q! \qquad \text{and} \qquad b(\lambda)=\prod_{j=1}^k[\lambda_j]_q.$$

Question.The following appears to be true. Is it? $$\prod_{\lambda\vdash n}a(\lambda)=\prod_{\lambda\vdash n}b(\lambda).$$

notfollow by transposing $\lambda$. However, it is equivalent toEnumerative Combinatorics, vol. 1, 2nd ed., Exercise 1.80, after conjugating $\lambda$. Note that the solution to this exercise gives another instance in which a proof was claimed to follow by conjugation. $\endgroup$ – Richard Stanley Oct 7 '18 at 18:48