For $n > 1$, show that every prime divisor of $n! + 1$ is an odd integer that is greater than $n$.
If $n > 1$, then $n! = n(n-1)(n-2) \cdots 3\cdot2\cdot1 = 2[n(n-1)(n-2) \cdots 3\cdot1]$ is even. So $n! + 1$ is odd. Does this mean every divisor of $n! + 1$ is odd? It's not like an even divisor can divide an odd number anyways.