The question is:
For all $n>2$, where $n \in \mathbb Z$: there exists $p$ prime such that $n<p<n!$
Here is my Proof:
$\forall$ $p<n: p|n!$, or $p$ divides $n!$
Since $n!$ and $n!-1$ are relatively prime
$=>$ $n!$ and $n!-1$ share no common divisors
$=>$ there must a prime $p > n$ such that $p | (n!-1)$
I feel like it needs to be clarified more. What else should I add? Thanks in advance!