Prove the existence or the non-existence of a couple of numbers ($n$,$m$) such that $n^2=m!$ In recent days, while I was doing exercises on combinatorics, I thought if a number $m!$ could be a perfect square. I proved to demonstrate it through the prime factorization. My attempt: $$k=m!=p_1^{a_1}\cdot p_2^{a_2}\cdot p_3^{a_3}....\cdot p_n^{a_n}$$
with $a_1,a_2,..,a_n$ that have to be even. Now if there exists a prime number $p_j$ between $\frac {k}{2}<p_j<k$ this prime number in the prime factorization the exponent ($a_j$) relative to the prime number $p_j$ is equal to $1$ ($a_j=1$) because the number $k$ isn't a multiple of $p_j$; for example $12!$ isn't a perfect square because the prime numbers $7$ and $11$ $6<7,11<12$ aren't divisors of $12$. 
My question
There exist a prime number $p_j$ between $k$ and $\frac {k}{2}$ and how can I prove if there exist or not exist a number $m!=n^2$?

Let $m\ge 3$ (otherwise $(m,n)=(0,\pm 1),(1,\pm 1)$).    
Assume for contr. $\ n^2=m!\ $ for some $n\in\Bbb Z$.   
Then $\ \exists p\in\Bbb P\ $ with $\ m/2<p<m\ $ by Berdrand's postulate.   
$\,\Rightarrow\, p\mid m!\,\Rightarrow\, p\mid n^2\,\Rightarrow\, p^2\mid n^2\,\Rightarrow p^2\mid m!$. But $m<2p$, contr.
