In some work I've been doing on the cohomology of the moduli space of curves, the following identity has come up:


$$\prod\_{i=1}^n \frac{x^{i-1}}{x^i-1} = \sum\_{(a\_1^{r\_1},\ldots,a\_{\ell}^{r\_{\ell}}) \vdash n} \left(\prod\_{j=1}^{\ell} \frac{1}{a\_i^{r\_i} (x^{a\_i}-1)^{r\_i} (r\_i)!}\right).$$


Here $x$ is a formal variable and the sum on the RHS is over all partitions of $n$. By $(a\_1^{r\_1},\ldots,a\_{\ell}^{r\_{\ell}}) \vdash n$, I mean a partition of the form


$$r\_1 a\_1 + r\_2 a\_2 + \cdots + r\_{\ell} a\_{\ell} = n$$


with $r\_1,\ldots,r\_{\ell} \geq 1$ and $a\_1>a\_2>\cdots>a\_{\ell} \geq 1$.


I have verified this identity with Mathematica for $1 \leq n \leq 20$. However, I cannot figure out how to prove that it is always true. Can anyone help me?


This reminds me a little bit of the identity in [this](https://mathoverflow.net/questions/224154/a-remarkable-sum-over-partitions) question, and I've tried without success to use the tools discussed in the answers to that question to solve it.




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EDIT: In case anyone is interested, a version of this identity now appears as Lemma 5.2 in my paper "The high dimensional cohomology of the moduli space of curves with level structures" (joint w/ Neil Fullarton), which can be downloaded from my webpage [here](http://www.nd.edu/~andyp/papers). Thanks to Lucia for telling me how to prove it!



Identity involving a sum over all partitions of $n$
Here's a quick sketch (since I'm pressed for time). Multiply both sides of the identity by $t^n$ and sum over $n$ from $0$ to infinity. From the cycle decomposition identity (Polya's formula) the right side becomes 
$$ 
\exp \Big( \sum\_{i=1}^{\infty} \frac{t^i}{i (x^i-1)} \Big)= \exp\Big( -\sum\_{i=1}^{\infty} \frac{t^i}{i} \sum\_{j=0}^{\infty} x^{ji} \Big) 
= \exp\Big( \sum\_{j=0}^{\infty} \log (1-t x^j) \Big) = \prod\_{j=0}^{\infty} (1-tx^j). 
$$ 
The RHS is also known as a Pochhammer symbol (see the Wikipedia article linked in my comment): it is $(t;x)\_{\infty}$. The wikipedia article already describes the combinatorial identity (simple partition relation) 
$$ 
(t;x)\_{\infty} = \sum\_{n=0}^{\infty} \frac{(-1)^n x^{n(n-1)/2} }{(x;x)\_n} t^n, 
$$ 
where 
$$ 
(x;x)\_n = (1-x) (1-x^2) \cdots (1-x^n). 
$$ 
This matches what you get from multiplying your LHS by $t^n$ and summing. 


