$n! \mid 2^{n\cdot (n+1)/2} \cdot \prod_{i=1}^{n} (2^i-1)$ for all $n \in \mathbb{N}$ I would like to prove that : 
$$n! \mid 2^{n\cdot(n+1)/2} \cdot \prod_{i=1}^{n} (2^i-1)$$
My attempt : 
I would like to prove that : $ v_p(n!) \leq v_p(\prod_{i=1}^{n} (2^i-1)) = \sum_{i=1}^n v_p(2^i-1)$
Then I tried to use Legendre theorem yet it doesn't seem to lead to anything...
 A: Note that $$\prod_{i=1}^n\,\left(2^n-2^{i-1}\right)=2^{\frac{n(n-1)}{2}}\,\prod_{i=1}^n\,\left(2^i-1\right)$$ is the number of ordered bases $\left(v_1,v_2,\ldots,v_n\right)$ of the $\mathbb{F}_2$-vector space $\mathbb{F}_2^n$.  If you want to compute the number of unordered bases $\left\{v_1,v_2,\ldots,v_n\right\}$, then you would need to ....
The same goes for any prime natural number $p$.  That is,
$$p^{\frac{n(n-1)}{2}}\,\prod_{i=1}^n\,\left(\frac{p^i-1}{p-1}\right)$$
is divisible by $n!$.  In fact, it is true that
$$p^{\left\lfloor\frac{n-1}{p-1}\right\rfloor}\,\prod_{i=1}^n\,\left(\frac{p^i-1}{p-1}\right)$$
is divisible by $n!$.
Even more generally, let $k$ be a positive integer.  Write $P_k$ for the set of prime natural numbers that divide $k$.  Then, $n!$ divides
$$k^{n(n-1)}\,\prod_{i=1}^n\,\prod_{p\in P_k}\,\left(\frac{1-1/p^i}{1-1/p}\right)\,.$$
A: For any prime $p$
$$v_p(n!) = \sum_{i=1}^\infty\left \lfloor{\frac{n}{p^i}}\right \rfloor  
\leq \left\lfloor{\sum_{i=1}^\infty\frac{n}{p^i}}\right\rfloor
=\left\lfloor{\frac{n}{p-1}}\right\rfloor$$
For any odd prime $p$, by Fermat's little theorem, $2^{p-1}\equiv 1 \mod p$, so $2^{(p-1)k}\equiv 1 \mod p$ for any $k\geq 1$, so $p|2^{(p-1)k}-1$, so 
$$p^{\left\lfloor{\frac{n}{p-1}}\right\rfloor}|\prod_{i=1}^n (2^i-1)$$
For $p=2$, $v_2(n!)\leq n$
In conclusion, $$n! \mid 2^{n} \cdot \prod_{i=1}^{n} (2^i-1)$$
I think your original problem of $n! \mid 2^{n\cdot(n+1)/2} \cdot \prod_{i=1}^{n} (2^i-1)$ is not sharp in the $2^{n\cdot(n+1)/2}$ part
