The question was stimulated by this one. Here it comes:
When you look at the sum $\sum\limits_{k=1}^N k!$ for $N\geq 10$, you'll always find $3$ and $11$ among the prime factors, due to the fact that $$ \sum\limits_{k=1}^{10}k!=3^2\times 11\times 40787. $$ Increasing $N$ will give rise to factors $3$ resp. $11$.
Are $3$ and $11$ the only common prime factors in $\sum\limits_{k=1}^N k!$ for $N\geq 10$?
I think, one has to show, that $\sum\limits_{k=1}^{N}k!$ has a factor of $N+1$, because the upcoming sum will always share the $N+1$ factor as well. This happens for $$ \underbrace{1!+2!}_{\color{blue}{3}}+\color{blue}{3}! \text{ and } \underbrace{1!+2!+\cdots+10!}_{3^2\times \color{red}{11}\times 40787}+\color{red}{11}! $$