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 $p$).