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Let $G$ be a group and consider for all $n$ the product $G^n=G\times\cdots\times G$. Define maps $\partial_i\colon G^n\to G^{n-1}$ by $$ \partial_i(g_1,\ldots,g_n) = \begin{cases} (g_2,\ldots,g_n), & \text{if } i=0, \\ (g_1,\ldots,g_i g_{i+1},\ldots,g_n) & \text{if } 0

This is quite easy to prove directly by case checking for different values of $k$. To get started, we may assume by symmetry that $k\neq 0$ and work from there. However, I suspect that there exists a beautiful proof that shows something fundamental about the associative law. In fact, I think the statement more or less is the associative law in some disguised form. Can someone see a pretty proof that does not rely on case-by-case checking?

Readers familiar with simplicial sets will see that I simply ask for a good argument why $\mathit{BG}_{\bullet}$ is a fibrant simplicial set. But the question is purely algebraic and I saw no reason to involve simplical sets in the formulation. Of course, answers involving simplicial methods are more than welcome.

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