Are $3$ and $11$ the only common prime factors in $\sum\limits_{k=1}^N k!$ for $N\geq 10$? 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 
$$
\underbrace{1!+2!}_{\color{blue}{3}}+\color{blue}{3}! \text{ and } \underbrace{1!+2!+\cdots+10!}_{3^2\times \color{red}{11}\times 40787}+\color{red}{11}!
$$

As pointed out in the comments, the case is trivial( at least if you know some theorems) if you fix $n>10$, instead of $n>p-1$ for prime $p$. 
It follows from Wilson's Theorem, that if you have a multiple of $p$ at index $n= p-2$ you won't at index $p-1$ because it will decrease out of being one for that index. $p>12$ implies at least $1$ index where it is NOT a multiple if it worked at index $11$. That leaves us with $p<12$ which would have to be factors of the sum up to $p-1$, $2$ is out as the sum is odd, $5$, needs $24+1+2+6=33$ to be a multiple of $5$, it isn't. Lastly $7$ needs $1+2+6+24+120+720=873$ to be a multiple which would force $33$ to be a multiple of $7$ which it isn't. 
