# Prove that any prime factor of $( x!+1)$ is larger than$x$.

I want to prove the statement "Any prime factor of $x!+1$ is larger than $x$."

Any slight hint will be ok.

Assume $p \leq x$ divides $x! + 1$. Then, $p$ also divides $x!$ (as any $y \leq x$ divides $x!$). Thus, $p$ divides both $x!$ and $x! + 1$. Do you see what this forces $p$ to be?