$(n!+1,(n+1)!)$ can be rewritten as $(n!+1,(n+1)*n!)$.
I know that if $n!$ is divisible by a prime $p$ then $p$ doesn't divide n!+1. So when I'm looking at then is $(n!+1 , n+1)$ which I can make $(n!-n,n+1)$ by subtracting $n+1$ from $n!+1$ and since gcd is preserved in linear combinations I still get the same gcd for $(n!-n,n+1)$. I then look at $n!-n = n[(n-1)!-1]$ again if a prime $p$ divides $n$ I'll find that $p$ doesn't divide $n+1$. So I'm looking at $((n-1)!-1,n+1)$. I've looked at the first few n's and it seems that the gcd is either 1 or n+1. But I'm stuck on how to get there from $((n-1)!-1,n+1)$.
Can anybody provide a hint as to how to proceed?