TITLE: $n!$ divides a product: Part I
QUESTION [14 upvotes]: Question. The following is always an integer. Is it not?
  $$\frac{(2^n-1)(2^n-2)(2^n-4)(2^n-8)\cdots(2^n-2^{n-1})}{n!}.$$

John Shareshian has supplied a cute proof. I'm encouraged to ask:

Question. Can you give alternative proofs, even if they are not particularly as short?

REPLY [6 votes]: Essentially the same argument: the ratio is a number of bases of the vector space $\mathbb{F}_2^n$.
A proof working also for composite values of $a=2$:
Let $p$ be a prime dividing $n!$. If $p$ divides $a$, then the product $\prod (a^n-a^k)$ is divisible by $p^{n-1}$, while $n!$ is not divisible by $p^n$. If $a$ and $p$ are coprime, it suffices to check that for any exponent $s$ the set $\{a^n-a^k\}$ contains at least as many numbers divisible by $p^s$ as the set $\{1,\dots,n\}$. It follows from the fact that a period of powers of a sequence $\{1,a,a^2,\dots\}$ modulo $p^s$ is less than $p^s$.