
> 
> **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?
> 
> 
> 



$n!$ divides a product: Part I
It is, because $S\_n$ embeds in $GL\_n({\mathbb F}\_2)$.


