Can $n!$ be a perfect square when $n$ is an integer greater than $1$? 
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.)

√n ≤ n/2 for n ≥ 4. Thus if p is a prime such that n/2 < p ≤ n, we have
√n < p → n < p² so p² cannot be a factor of n if n ≥ 4.
