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.

Find the exponent of the largest power of 2 that divides both sides. In RHS it is $0 + 1 + \ldots + (n-1) = \frac{(n-1)n}{2}$. In LHS it can be found with Legendre's formula, which gives $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor$. Since $\sum_{i=1}^{\infty} \lfloor \frac{k}{2^i} \rfloor < \sum_{i=1}^{\infty} \frac{k}{2^i} = k$ (the inequality is strict because at least one term in the bracket is not an integer), we have $k > \frac{(n-1)n}{2}$, or $k \ge \frac{(n-1)n}{2}+1$.
Since $RHS < 2^{n^2}$, and factorial grows faster than any exponential function, for large values of $n$, LHS will be larger than RHS. We need to find out an upper bound for $n$.
When $n \ge 6$, $$k! \ge (\frac{(n-1)n}{2}+1)! \ge 7! \cdot 8^{\frac{(n-1)n}{2}-6} > 2^{12} \cdot 2^{\frac{3}{2}n^2 - \frac{3}{2}n - 18} = 2^{n^2} \cdot 2^{\frac{1}{2}n^2 - \frac{3}{2}n - 6}$$
And $\frac{1}{2}n^2 - \frac{3}{2}n - 6 > 0$, so LHS > RHS. Therefore there are no solutions with $n \ge 6$. Manually checking the remaining cases gives the only solutions $(1, 1)$ and $(3, 2)$.
