Math Help - Gcd problem

1. Gcd problem

Proof that:

gcd (n!+1 , (n+1)! + 1) = 1

2. Suppose that there exists a $p \geq 2$ such that $p \mid n!+1$. If $p \mid a$ then $p \mid ca$ where c is an integer. Using this $p \mid (n+1)(n!+1)$. Now look at $(n+1)!+1$. If $p \mid (n+1)!+1=(n+1)(n!+1)-n$ we come to the conclussion that $p \mid -n$ but this is impossible since $n!+1 \equiv 1 \pmod n$ so p=1 and this is a contradiction so there exist no number p such that p divides $n!+1$ and $(n+1)!+1$ and it follows that $GCD(n!+1,(n+1)!+1)=1$