Divisibility and Lagrange Theorem

110 Views Asked by At

I came across the following question from math olympiad:

For $n\in \mathbb{Z}^+$, prove that $$n!\mid\prod_{k=1}^n(2^n-2^{k-1}).$$

While I can solve it using elementary number theory, I notice that $\prod_{k=1}^n(2^n-2^{k-1})$ is the order of the general linear group $GL_n(\mathbb{F}_{2})$ over $\mathbb{F}_{2}$ and $n!$ is the order of the symmetric group $S_n$.

So I was wondering can we view $S_n$ as a subgroup of $GL_n(\mathbb{F}_{2})$ and invoke Lagrange theorem to solve the question, thank you so much.