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Let $\lambda=(\lambda_1,...,\lambda_r,...)$ be a partition (i.e. $\lambda_i\ge \lambda_{i+1}$ and there are only finitely many non-zero terms.) Let $\lambda'$ be a conjugate partition, i.e. $\lambda_i'=\mathrm{card}\{j:\ \lambda_j\ge i\}$.

Let $\Lambda:=\{(i,j)\in\mathbb{Z}^2 :\ 1\le j\le \lambda_i \}$ be the diagram of $\lambda$.

Take $x=(i,j)$ and define the hook-length by $$h(x)=\lambda_i+\lambda_j'-i-j+1.$$ From the properties of diagrams it is known that for $$f_{\lambda,n}(t):=\sum\limits_{i=1}^nt^{\lambda_i+n-i}$$ with $n\ge \lambda_1', \ m\ge\lambda_1$ the following identity holds $$f_{\lambda,n}(t)+t^{m+n-1}f_{\lambda',m}(t^{-1})=\frac{1-t^{n+m}}{1-t}.$$ From this it follows (after some calculations) that $$\prod\limits_{x\in\Lambda}(1-t^{h(x)})\prod\limits_{i<j}(1-t^{\mu_i-\mu_j})=\prod\limits_{i\ge 1}\prod\limits_{j=1}^{\mu_i}(1-t^j),$$ where $\mu_i=\lambda_1+n-i$.

I would like to show that the above relation implies $$\prod\limits_{x\in\Lambda}h(x)\prod\limits_{i<j}(\mu_i-\mu_j)=\prod\limits_{i\ge 1}\mu_i!.$$

I know that I should somehow divide the previous relation by $(1-t)^{|\lambda|}$, but I don't see how to use this hint in a way that is not so tedious.

mikis
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    The idea is to divide by an appropriate power of $1-t$, and then `evaluate' at $t=1$ - the point is that $(1-t^r)/(1-t)=1+t+\cdots+t^{r-1} = r$ at $t=1$. – Matthew Towers Feb 27 '19 at 18:37
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    The power series $1-t^1, 1-t^2, 1-t^3, \ldots$ are multiplicatively $\mathbb{Z}$-linerarly independent. That is, if $a_1, a_2, a_3, \ldots$ are integers such that all but finitely many $i \geq 1$ satisfy $a_i = 0$ and such that $\prod\limits_{i \geq 1} \left(1-t^i\right)^{a_i} = 1$, then $a_1 = a_2 = a_3 = \cdots = 0$. (Actually, this holds even without the "all but finitely many $i \geq 1$ satisfy $a_i = 0$" condition.) Thus, the equality $\prod\limits_{x\in\Lambda}(1-t^{h(x)})\prod\limits_{i<j}(1-t^{\mu_i-\mu_j})=\prod\limits_{i\ge 1}\prod\limits_{j=1}^{\mu_i}(1-t^j)$ yields ... – darij grinberg Feb 27 '19 at 18:45
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    ... that the multiset of the exponents of $t$ appearing on the left hand side equals the multiset of the exponents of $t$ appearing on the right hand side. – darij grinberg Feb 27 '19 at 18:46
  • Thank you so much! Actually, I was thinking about that, but did some stupid mistake in the calculations... – mikis Feb 27 '19 at 18:53

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