True or False: For $n>1$, $n!$ can never be a perfect square. I am trying to solve the following: 
True or False: For $n>1$, $n!$ can never be a perfect square.
I am thinking on the following lines:
Any perfect square $N$ is of the form $N=p_1^{2m_1}p_2^{2m_2}p_3^{2m_3}\ldots p_k^{2m_k}$ where $p_1,p_2,\ldots,p_k$ are the prime factors of $N$ and $m_1,m_2,\ldots, m_k$ are positive integers. So, using elementary combinatorics, the number of divisors of $N$ is $(2m_1+1)(2m_2+1)\ldots(2m_k+1)$, which is odd. So, any perfect square has odd number of divisors. Now if we can show that $n!$ has even number of divisors, then it can be concluded that $n!$ is not a perfect square. But I am stuck at this point. Can someone help me out? Or are there better ways to proceed ?
 A: I think I have an answer now. Thanks to the comment by JimmyK4542.
If $n$ is prime, then its obvious that $n!$ is not a perfect square, for in that case $n\mid n!$ but $n^2\nmid n!$. If $n$ is not a prime, then let the $r$th prime $p_r$ be the largest prime which is less than $n$. Now, $n!=1\cdot 2\cdot 3\cdot\ldots \cdot p_r\cdot (p_r+1)\cdot (p_r+2)\cdot \ldots \cdot (p_r+n-p_r)$. We can show that $p_r$ does not divide any of the numbers $p_r+1, p_r+2,\ldots,n$. This is because any number greater than $p_r$ which is  divisible by $p_r$ must be of the form $mp_r$ where $m\geq 2$. But by Bertrand's postulate, $p_{r+1}<2p_r$, where $p_{r+1}$ is the $(r+1)$th prime. Thus if any of the numbers $p_r+1, p_r+2,\ldots,n$ is divisible by $p_r$ then one of these numbers must be the $(r+1)$th prime. But this contradicts the fact that $p_r$ is the largest prime which is less than $n$. So none of the numbers $p_r+1, p_r+2,\ldots,n$ is divisible by $p_r$. So, we see that $p_r\mid n!$ but $p_r^2\nmid n!$. Hence, $n!$ is not a perfect square. This also proves that $n!$ actually has even number of divisors.
I shall appreciate any further suggestion fo rthe improvement of this answer.
