For all $n>2$ there exists a prime number between $n$ and $ n!$ How to prove that there exists a prime number between $n$ and $ n!$, for all $ n> 2$?

(Bertrand's postulate gives a much better bound, but this question is about obtaining a self-contained proof.)

For $n=2$ take $p=2$. For $n\geq 3$, take $p$ as a prime divisor of $n!-1$ (hence $p<n!$). Then $p>n$, if not, $p$ divide $n!$, hence divide $n!-(n!-1)=1$. 
