Can $n!$ be a perfect square when $n$ is an integer greater than $1$?
Clearly, when $n$ is prime, $n!$ is not a perfect square because the exponent of $n$ in $n!$ is $1$. The same goes when $n-1$ is prime, by considering the exponent of $n-1$.
What is the answer for a general value of $n$? (And is it possible, to prove without Bertrand's postulate. Because Bertrands postulate is quite a strong result.)
Best Answer
Assume, $n\geq 4$. By Bertrand's postulate there is a prime, let's call it $p$ such that $\frac{n}{2}<p<n$ . Suppose, $p^2$ divides $n$. Then, there should be another number $m$ such that $p<m\leq n$ such that $p$ divides $m$. So, $\frac{m}{p}\geq 2$, then, $m\geq 2p > n$. This is a contradiction. So, $p$ divides $n!$ but $p^2$ does not. So, $n!$ is not a perfect square.
Bertrand's postulate
That leaves two more cases. We check directly that, $2!=2$ and $3!=6$ are not perfect squares.