Proving gcd(nn!, n+1)=1 using prime factorization

[youvecaughtme]

Homework Statement

For any n \in \mathbb{N}, find \mathrm{gcd}(n!+1,(n+1)!+1). First come up with a conjecture, then prove it.

2. The attempt at a solution
By testing some values, it seems like \mathrm{gcd}(n!+1,(n+1)!+1) = 1

I'm trying to prove this by induction. I'll leave out the inductive assumption and base case verification because I can do those.

I have \mathrm{gcd}(n!+1,(n+1)!+1) = 1 and I'm trying to show that \mathrm{gcd}((n+1)!+1,(n+2)!+1) = 1.

I can simplify what's given to me to \mathrm{gcd}(nn!, n!+1)=1 but I can't find out how to get it into the form I want it. Can anybody look at what I'm doing and give me any guidance?

\mathrm{gcd}(n!+1,(n+1)!+1) = 1 \implies \mathrm{gcd}(n!+1,(n+1)n!+1) = 1 \implies \mathrm{gcd}(n!+1,nn!+n!+1) = 1 \implies \mathrm{gcd}((n)n!, n!+1) = 1

 

[morphism]

I wouldn't bother with induction here.

Try to argue as follows. If a prime divides both n!+1 and (n+1)!+1, it will divide their difference. What does this imply?