This question is inspired from @Mathphile's problem:
The value$\sum_{i=1}^n i!$ where $n \in \mathbb{N}$, is only semiprime for $n=3,4$
One can easily solve this conjecture by knowing that $9 \mid \sum_{i=1}^8 i!$ which implies that $9 \mid \sum_{i=1}^n i!$ for any $n \geqslant 8$. We can notice: $$p \mid \sum_{i=1}^{p-1}i! \implies p \mid \sum_{i=1}^ni! \quad (n \geqslant p-1)$$ Are there finitely or infinitely many primes of this form? If there are an infinite number of them (which heuristically seems to be the case), we consequently have $\sum_{i=1}^n i!$ to have an arbitrarily large number of divisors (or prime divisors) $\forall$ $n>N$ for sufficiently large $N$.
Can we extend this to $p^k \mid \sum_{i=1}^{p^k-1} i!$ (for this to hold, it must hold for all powers $p^j$ for $1 \leqslant j \leqslant k$)?