$\gcd(n!+1,(n+1)!)$ The recent post didn't really provide sufficient help.  It was too vague, most of it went over my head.
Anyway, I'm trying to find the $\gcd(n!+1,(n+1)!)$.
First I let $d=ab\mid(n!+1)$ and $d=ab\mid(n+1)n!$ where $d=ab$ is the GCD.
From $ab\mid(n+1)n!$ I get $a\mid(n+1)$ and $b|n!$.
Because $b\mid n!$ and $ab\mid(n!+1)$, $b$ must be 1.
Consequently, $a\mid(n!+1)$ and $a\mid(n+1)$.
So narrowing down options for $a$ should get me an answer.  At this point I've tried to somehow bring it around and relate it to Wilson's theorem as this problem is from that section of my textbook, but I seem to be missing something.  This is part of independent study, though help of any kind is appreciated.

$$
\begin{align}
\gcd(n!+1,(n+1)!)
&=\gcd(n!+1,n!)\gcd(n!+1,n+1)\\
&=\gcd(n!+1,n+1)
\end{align}
$$
Wilson's theorem says that $n!+1=0\pmod{n+1}$ iff $n+1$ is prime. Therefore, if $n+1$ is prime, $\gcd(n!+1,n+1)=n+1$.
Suppose $n+1$ is composite. Let $p\,|\,n+1$ where $p<n+1$ is prime. Since $p\le n$, $p\,|\,n!$ Therefore, $p\not|\,n!+1$. This means that $\gcd(n!+1,n+1)=1$.
In conclusion, we get that
$$
\gcd(n!+1,(n+1)!)=\left\{\begin{array}{}n+1&\text{if }n+1\text{ is prime}\\1&\text{if }n+1\text{ is not prime}\end{array}\right.
$$
