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

Proof. Let $d$ be a common divisor. Then $n!\equiv-1\pmod d$. Since $(n+1)!=(n+1)\,n!$,
$$0\equiv(n+1)!+1\equiv-(n+1)+1=-n\pmod d,$$
so $d\mid n$. But $n\mid n!$, hence $d\mid n!$. Together with $d\mid n!+1$ this gives $d\mid1$.

Check. $n=3$: $\gcd(7,25)=1$; $n=4$: $\gcd(25,121)=1$ ✓.

Remark. Equivalently, $\gcd(n!+1,(n+1)!+1)=\gcd(n!+1,n)=1$ by the Euclidean algorithm.

Lesson. To prove coprimality, assume a common divisor $d$, reduce one expression modulo $d$ using the other, and show that $d$ divides $1$.