TITLE: Hint for $(n!+1,(n+1)!)$, stuck at $ (n!+1,n+1)$
QUESTION [10 upvotes]: $(n!+1,(n+1)!)$ 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?

REPLY [2 votes]: Let
$gcd(n!+1,(n+1)!)=d$.\
$\therefore d/n!+1, d/(n+1)n!$.\
$\Rightarrow d/n!+1,d/n+1,d/n!$.\
$\Rightarrow d/n!+1-n!$.\
Hence we get $d/1$.
It follows that $d=1$.