I guess the canonical proof to this problem is attained using the Euclidean Algorithm (I've seen some posts like this already). I came with this proof, based on gcd definition and some divisibility properties. I'd like to know if it seems correct to you.

**Proof**:

Let $d$ be a common divisor of both terms. Then $n! +1$ = $d$.$s$ and $(n+1)! +1 = d.t$, for $s, t$ in $\Bbb Z/{0}$. Therefore $(n +1)n! + 1 = n.n! + n! + 1 = n.n! + d.s = d.t$, and it follows that $d|n.n!$. Now, since $n=d(t-s)/n!$, $d|n$. Then $d$ must divide 1 ($n! + 1 = d.s$), and $|d|$ = 1. Being $d'$ another common divisor, it is clear that $d'$ must divide 1 also, and therefore it does divide $d$, so we have $d = gcd(n! + 1, (n+1)! +1) = 1$