TITLE: Second part of the factorial sum divisibility question
QUESTION [48 upvotes]: Which primes $p$ divide the sum of factorials $1! + 2! + 3! + 4! + 5! + \cdots + (p-1)!$? This is related to my previous question.

REPLY [2 votes]: This is what I thought about, it should be some comment, but this exceeds the character limit for comment.
If you take Wilson Theorem that 
$(p-1)!\equiv -1 \mod{p}$, then
$(p-2)!\equiv 1 \mod{p}$.
If we take inverse of $a$ in $F_p$ as $\dfrac{1}{a}$, we know that $\dfrac{1}{a} + \dfrac{1}{b} = \dfrac{a+b}{ab}$ in $F_p$, it can taken as usual addition.
Then 
$(p-3)! \equiv -\dfrac{1}{2}\mod{p}$
$(p-4)! \equiv -\dfrac{1}{2}\cdot\left(-\dfrac{1}{3}\right)\mod{p}$ 
$\cdots$
$1! \equiv (-1)^{p-3}\dfrac{1}{(p-2)!} \mod{p}$
Thus 
$1!+2!+\cdots(p-3)! \equiv -\sum_{k=0}^{p-2}(-1)^k\dfrac{1}{k!}\mod{p}$
And we state without proof.
Lemma:
Consider the fraction $\dfrac{p}{q}$ with $q = m!$, then the closest number to $\exp(-1)$ with form $\dfrac{p}{q}$ should satisfy
$$\dfrac{p}{q} = \sum_{j=0}^m (-1)^j\dfrac{1}{j!}$$
Then 
$-\sum_{k=0}^{p-2}(-1)^k\dfrac{1}{k!} \equiv - \mathrm{floor}\left[\dfrac{(p-2)!}{\exp(1)}\right]\mod{p}$
However, this cannot be checked for large $p$, since the constant $\exp(1)$ is not accurate enough. I simply checked for smaller primes.
However, I felt that there should be other solutions than 3 and 11, otherwise, this can be a very interesting property for the constant.