There are many proofs for Cauchy's Theorem from group theory, which states that if a prime $p$ divides the order of a finite group $G$, then $\exists g\in G$ of order $p$.

Recently I've encountered an elegant combinatorial proof for this theorem *in the abelian case*. It is self-contained, and appears in Hecke's "Lectures on the Theory of Algebraic Numbers". It has 3 steps:

- Assume $p \mid |G|$, where $G=\{g_1,\cdots, g_n\}$. Consider all the product $\prod_{i=1}^{n} g_i^{a_i}$ where $0 \le a_i < \text{ord}(g_i)$.
- Each element of $G$ appears in this multiset of size $\prod_{i=1}^{n} \text{ord}(g_i)$ an equal amount of times. This is the (only) point where the fact that $G$ is abelian is used.
- This implies $\prod_{i=1}^{n}\text{ord}(g_i)$ is divisibly by $p$. As $p$ is prime, we deduce $p \mid \text{ord}(g_i)$ for some $1\le i \le n$. Now we take a suitable power of $g_i$. $\blacksquare$

My question is: Can the second step be generalized to the general case? Is it true that the multiset of products $\{ \prod_{i=1}^{n} g_i^{a_i} \mid 0 \le a_i < \text{ord}(g_i) \}$ contains each element of $G$ the same amount of times? Note that there's no "canonical" product $\prod_{i=1}^{n} g_i^{a_i}$ in the non-abelian case - there are several ways to interpret this product (maybe the $g_i$'s come before the $g_j$'s for $i<j$; maybe they interlace), and I don't want to specify the way.

I've read a bit about equations in groups, and haven't seen this kind of exponential problem discussed.

(I am also interested in the origins of this proof, by the way. If some one knows, please drop a comment.)