Is $\sqrt{n!}$ a natural number?

Solution 1:

For any $n \gt 1$ there will be some prime in the range $(n/2,n]$ which will only occur once in the factorization of $n!$ by Bertrand's Postulate. This will ensure that $\sqrt{n!}$ is not an integer.

Solution 2:

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.