TITLE: Identity involving a sum over all partitions of $n$
QUESTION [7 upvotes]: 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 question, and I've tried without success to use the tools discussed in the answers to that question to solve it.

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.  Thanks to Lucia for telling me how to prove it!

REPLY [15 votes]: 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.