$\prod_{i=0}^{n-1}(2^n-2^i)$ can be divided by $n!$

A nice solution is given by the natural embedding of $S_n$ into $\mathrm{GL}_n(\mathbb{F}_2)$.

Here $S_n$ is the symmetric group of $n$ elements. Every element of $S_n$ permutes the canonical basis of $\mathbb{F}_2^n$, hence gives an $\mathbb{F}_2$-linear automorphism of $\mathbb{F}_2^n$. This embeds $S_n$ as a subgroup of $\mathrm{GL}_n(\mathbb{F}_2)$.

Now it only remains to say that the cardinalities of the two groups are $n!$ and $\prod_{i = 0}^{n - 1}(2^n - 2^i)$, respectively.

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By replacing $\mathbb{F}_2$ with $\mathbb{F}_q$ for any prime power $q$, one proves in the same way that $n!$ divides $\prod_{i = 0}^{n - 1}(q^n - q^i)$.

Hint For each odd prime $p$, if $p^k |m \leq n$ then $$p^k |2^{\frac{p-1}{p}m}-1$$ and hence $$p^k| 2^{n}-2^{n-\frac{p-1}{p}m}$$