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.)
 A: Hopefully this is a little more intuitive (although quite a bit longer) than the other answers up here. 
Let's begin by stating a simple fact : (1) when factored into its prime factorization, any perfect square will have an even number of each prime factor.
If $n$ is a prime number, then $n$ will not repeat in any of the other factors of $n!$, meaning that $n!$ cannot be a perfect square (1). Consider if $n$ is composite. $n!$ will contain at least two prime factors ($n=4$ is the smallest composite number that qualifies the restraints), so let's call $p$ the largest prime factor of $n!$ 
The only way that $n!$ can be a perfect square is if $n!$ contains $p$ and a second multiple of $p$ (1). Obviously, this multiple must be greater than $p$ and less than $n.$ 
Using Bertrand's postulate, we know that there exists an additional prime number, let's say $p'$, such that $p < p' < 2p$. Because $p$ is the largest prime factor of $n!$, we know that $p' > n$ (If it were the opposite, then we would reach a contradiction). 
Thus it follows that $2p > p' > n$. Because $2p$ is the smallest multiple of $p$ and $2p > n$, then $n!$ only contains one factor of $p$. Therefore it is impossible for $n!$ to be a perfect square.
A: *

*If $n$ is prime, then for $n!$ to be a perfect square, one of $n-1, n-2, ... , 2$ must contain n as a factor. But this means one of $n-1, n-2, ... , 2 \geq n$, which is impossible. 

*If $n$ is not prime, then the first prime less than $n$ will be $p = n-k$, $0<k<n-1, 2\leq p<n$. No number less than $p$ will contain $p$ as a factor, so for $n!$ to be a perfect square there must exist a multiple of $p$, I'll call it $bp$, $1<b<n,$ such that$ p<bp\leq n$. 
Now according to chebyshev's theorem for any no. $p$ there exists a prime number between $p$and $2p.$ 
so if $r< n < 2r$ and also $p<n$ , so such an $n!$ would never be a perfect square. Hope this helps.
You can refer this. 
A: 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.
A: Your statement has a generalization.  There is a work by Erdos and Selfridge stating that the product of at least two consecutive natural numbers is never a power.  Here is it: https://projecteuclid.org/euclid.ijm/1256050816.
A: There is a prime between n/2 and n, if I am not mistaken. 
A: √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.
