Let $s\_n = \sum\_{i=1}^{n-1} i!$ and let $g\_n = \gcd (s\_n, n!)$. Then it is easy to see that $g\_n$ divides $g\_{n+1}$. The first few values of $g\_n$, starting at $n=2$ are $1, 3, 3, 3, 9, 9, 9, 9, 9, 99$, where $g\_{11}=99$. Then $g\_n=99$ for $11\leq n\leq 100,000$.


Note that if $n$ divides $s\_n$, then $n$ divides $g\_m$ for all $m\geq n$. If $n$ does not divide $s\_n$, then $n$ does not divide $s\_m$ for any $m\geq n$.


If $p$ is a prime dividing $g\_n$ but not dividing $g\_{n-1}$ then $p=n$, for if $p<n$ then $p$ divides $(n-1)!$ and therefore $p$ divides $s\_n-(n-1)!=s\_{n-1}$, whence $p$ divides $g\_{n-1}$.


So to show that $g\_n\rightarrow \infty$ it suffices to show that there are infinitely many primes $p$ such that $1!+2!+\cdots +(p-1)! \equiv 0$ (mod $p$).



What is the limit of gcd(1! + 2! + ... + (n-1)! ,  n!) ?
This is so close to the Kurepa conjecture which asserts that $\gcd\left(\sum\_{k=0}^{n-1}k!,n!\right)=2$ for all $n\geq 2$, which was settled in 2004 by D. Barsky and B. Benzaghou "Nombres de Bell et somme de factorielles". So what they proved is that $K(p)=1!+\cdots+(p-1)!\neq -1\pmod{p}$ for any odd prime $p$. This goes against Kevin Buzzard's heuristic that $K(p)$ is random mod $p$. Let me mention two ways you can restate the fact $p|K(p)$:


a) It is equivalent to $K(\infty)=\sum\_{k=1}^{\infty}k!$ not being a unit in $\mathbb Z\_p$.


b) It is equivalent to $\mathcal B\_{p-1}=2\pmod{p}$ where $\mathcal{B} \_n$ is the $n$th Bell number. (It is easy to show that $\mathcal B \_{p}=2\pmod{p}$)


I forgot to mention that the conjecture that $p>11$ doesn't divide $K(p)$ is in question B44 of R. Guy's "Unsolved Problems in Number theory".


