While studying some seemingly unrelated topological questions, I have experimentally discovered what appears (to me) to be a remarkable sum over partitions. I was wondering if anyone knows how to prove it.


Fixing $n \geq 1$, it can be stated as follows:


$$1=\sum\_{(a\_1^{k\_1},\ldots,a\_p^{k\_p}) \vdash n} \left(\frac{1}{(a\_1^{k\_1}) (k\_1 !)}\right) \left(\frac{1}{(a\_2^{k\_2}) (k\_2 !)}\right)\cdots\left(\frac{1}{(a\_p^{k\_p}) (k\_p !)}\right).$$
Here the sum is over all unordered partitions of $n$, and the symbol $(a\_1^{k\_1},\ldots,a\_p^{k\_p})$ denotes the partition where $a\_1$ appears $k\_1 \geq 1$ times, where $a\_2$ appears $k\_2 \geq 1$ times, etc, and where $a\_1 > a\_2 > \cdots > a\_p > 0$.


I have numerically verified this up to $n=6$.



A remarkable sum over partitions
Here is a more informative version of this identity. Let $Z\_n$ denote the *cycle index polynomial* of the symmetric group $S\_n$, namely


$$Z\_n = \frac{1}{n!} \sum\_{\sigma \in S\_n} z\_1^{c\_1(\sigma)} z\_2^{c\_2(\sigma)} \dots $$


where $c\_i(\sigma)$ denotes the number of $i$-cycles of $\sigma$. The observations in the comments boil down to the observation that


$$Z\_n = \sum\_{\sum ic\_i = n} \prod\_{i=1}^n \frac{z\_i^{c\_i}}{i^{c\_i} c\_i!}$$


and setting all $z\_i = 1$ gives your identity, but without doing this, it is possible to arrange the $Z\_n$ into a beautiful generating function, namely


$$\sum\_{n \ge 0} Z\_n t^n = \exp \left( \sum\_{n \ge 1} \frac{z\_i t^i}{i} \right).$$


This is the "permutation form" of the [exponential formula](https://en.wikipedia.org/wiki/Exponential_formula), which has many applications in combinatorics. See, for example, Stanley's *Enumerative Combinatorics: Volume 2* for an exposition, or [this blog post](https://qchu.wordpress.com/2009/06/24/gila-vi-the-cycle-index-polynomials-of-the-symmetric-groups/).


