Find all pairs $(k, n)$ of positive integers such that $k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$ Find all pairs $(k, n)$ of positive integers such that
$$k! = (2^n − 1)(2^n − 2)(2^n − 4) · · · (2^n − 2^{n−1})$$
I tried to solve this problem but only found one solution $(1,1)$. Please help me to solve this problem.

Via Legendre's formula, we have for any prime number $p$
$$ \nu_p(k!)=\sum_{i=1}^{\infty}\left \lfloor \frac{k}{p^i} \right \rfloor
\le \sum_{i=1}^{\infty} \frac{k}{p^i} =\frac{k}{p-1} $$
Note that $\nu_2(2^n-2^i)=i$, and hence
$$\nu_2(k!)=\nu_2\left(\prod_{i=0}^{n-1}({2^n-2^i})\right)=\sum_{i=0}^{n-1}i=\frac{n(n-1)}{2}$$
Thus 
$$\frac{n(n-1)}{2}\le k \tag{1}$$
Note that $\nu_3(2^n-2^i)=\nu_3(2^{n-i}-1)$ and 
$3\nmid 2^m-1$ if $m$ is odd. In addition $$\nu_3(2^{2m}-1)=\nu_3(m)+1$$
which is a corollary of lifting the exponent lemma.
Thus
$$\nu_3(k!)=\nu_3\left(\prod_{i=0}^{n-1}({2^n-2^i})\right)
= \left [ \frac{n}{2} \right] + \sum_{i=1}^{\left [ \frac{n}{2} \right]} \nu_3(i) 
=\nu_3 (\left[ \frac{n}{2} \right]!) + \left [ \frac{n}{2} \right]
\le \frac{\frac{n}{2}}{2} +\frac{n}{2} =\frac{3n}{4}$$
Note that $\nu_3(k!)\ge \left[ \frac{k}{3} \right] >\frac{k}{3}-1$. Thus $$\frac{k}{3}-1<\frac{3n}{4} \tag{2}$$
From $(1)$ and $(2)$ we know $n \in \{1,2,3,4,5,6\}$. Via verification we get the solutions are $(k,n)=(1,1)$ or $(3,2)$
