TITLE: $n! \mid 2^{n\cdot (n+1)/2} \cdot \prod_{i=1}^{n} (2^i-1)$ for all $n \in \mathbb{N}$
QUESTION [1 upvotes]: 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...

REPLY [1 votes]: 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
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