show $k! = m^\alpha$ is impossible I am not sure how to show that $$m^\alpha = k!$$ is impossible $\forall m,\alpha,k > 1$. I am supposed to use Bertrands theorem which states:  There exists atleast one prime in the open interval $(x, 2x)$ This is a homework question.

Note that since $a>1$, every prime number appearing in the decomposition of $m^a$ has power greater than 1.
Let $k>2$. If $k$ is even,  there is a prime number $p$ between $k$ and $k/2$ and this prime number $p$ appears only once in $k!$, hence $k!\neq m^a$.
When $k$ is odd, there is a prime nubmer $p$ between $k-1$ and $(k-1)/2$  and this prime number $p$ appears only once in $k!$, hence $k!\neq m^a$.
This is a basic idea and you need to fill gaps.
