TITLE: Is $\sqrt{n!}$ a natural number?
QUESTION [29 upvotes]: I'm new here (on Mathematics Stack Exchange).
Also, I'm a 10th grade student not a math expert.
So, my question is whether, 
$$\sqrt {n!} $$
comes in the set of the Natural Numbers. 
There were some answers when I tried to search on Google. But, it was out of my knowledge, that is, I didn't read/learnt the methods, functions, etc. they were using. So, if possible give a detailed explanation.
And pardon me if there is some good answer there and if if it's there then please share a link.

REPLY [17 votes]: Look at the prime factors of $n!$. If the square root of $n!$ was an integer, then $n!$ would be the square of an integer, and in the square of an integer, all prime factors occur an even number of times.
For example, if you take $100!$, which ends with $\cdots 95\times 96\times 97\times 98\times 99\times 100$, you see the prime number $97$. That prime number only occurs once in the factorisation of $100!$. All primes from $51$ to $100$ occur only once in the factorisation of $100!$. Therefore $100!$ is not a square. 
There is a theorem that there is always a prime number between $n$ and $2n$, and therefore any factorial starting with $2$! has one prime factor that only comes up once.