Easy. We just need to show that there do not exist $n, m$ such that $n! = 2^mm!$ and $m \ge 2.$ If $n! = 2^mm!$ then we have $(n-m+1)(n-m+2)...(n-1)n = 2^m$. Note that this is impossible for $n > 2$ because we will have at least one odd factor greater than $1$ in the left hand side, while the right side is a power of $2$. However, if $n \le 2$ we have no solutions at all. Thus there are no such $n, m$.