More Proof

First let's rewrite $\prod_{k=0}^{n-1} (2^n - 2^k) = \prod_{k=0}^{n-1} 2^k(2^{n-k} - 1) = 2^{t_{n-1}} \prod_{k=0}^{n-1} (2^{n-k}-1) = 2^{t_{n-1}} \prod_{k=1}^{n} (2^k - 1)$ where $t_n$ are the triangular numbers.

Now, according to De Polignac's Formula the highest power of a prime $p$ that divides $n!$ is $\sum_{k=1}^{\infty} \lfloor \frac{n}{p^k} \rfloor \leq \lfloor \sum_{k=1}^{\infty} \frac{n}{p^k} \rfloor = \lfloor \frac{n}{p-1} \rfloor$ . For $p=2$ this is $n$ which is smaller than or equal to $t_{n-1}$ whenever $n>2$ so our product has a high enough power of $2$ whenever $n>2$ . (And the cases n=1,2 can be checked to hold by computation).

For odd primes simply apply Fermat's little theorem which says that $2^{p-1} - 1 \equiv 0 \mod p$ to find that $\prod_{k=1}^{n} (2^k - 1)$ will be divided at least once by $p$ for each multiple of $p-1$ smaller than or equal to $n$ so this entire product is divided by $p^{\lfloor \frac{n}{p-1} \rfloor}$ . This means that indeed $\prod_{k=0}^{n-1} (2^n - 2^k)$ is divisible by $n!$ .