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

By testing some values, it seems like $\gcd(n!+1,(n+1)!+1) = 1$

I can simplify what's given to me to $\gcd(nn!, n!+1)=1$ 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?

$\gcd(n!+1,(n+1)!+1) = 1 \implies \gcd(n!+1,(n+1)n!+1) = 1 \implies \gcd(n!+1,nn!+n!+1) = 1 \implies \gcd(nn!, n!+1) = 1$
You don’t need to use induction; you just need to prove the statement in the title. Suppose that $p$ is a prime factor of $nn!$; can $p$ divide $n!+1$?
I dont know i am right or wrong but i can do this example in following way,\
Let $$\gcd(n\cdot n!,n!+1)=d$$
    $$\therefore d\mid n\cdot n!\ ,\ d\mid n!+1$$
    $$\Rightarrow d\mid n,\ d\mid n!,\ d\mid n!+1$$
    $$ \Rightarrow d\mid n!+1-n!$$
Thus $d\mid 1\ \Rightarrow d=1.$