Is the finite sum of factorials constant modulo the summation limit? The answer to the following question would give an alternative solution to an old olympiad question if it is true.
Prove that there is no (constant) integer $c$ such that
$$1!+2!+\dots + q! \equiv c \bmod q \text{ for all $q \in \mathbb N^\ast$.}$$
($\mathbb N^\ast = \mathbb N \setminus \{0\}$)
 A: Let $ K ( q ) = \sum _ { k = 1 } ^ { q - 1 } k ! $. Since $ q ! \equiv 0 \pmod q $, thus $ c $ is an integer such that for every positive integer $ q $, we have $ K ( q ) \equiv c \pmod q $. for instance, we have $ K ( q ! ) \equiv c \pmod { q ! } $. But by definition of $ K $, we know that $ K ( q ! ) \equiv K ( q ) \pmod { q ! } $, which leads to $ K ( q ) \equiv c \pmod { q ! } $. Hence there is a sequence of integers like $ ( k _ q ) _ { q \in \mathbb Z ^ + } $ such that $ c = k _ q \cdot q ! + K ( q ) $. Now for every positive integer $ q $:
$$ 0 = c - c = k _ { q + 1 } \cdot ( q + 1 ) ! + K ( q + 1 ) - k _ q \cdot q ! - K ( q ) = \big( ( q + 1 ) k _ { q + 1 } - k _  q + 1 \big) q ! $$
$$ \therefore \quad k _ { q + 1 } = \frac { k _ q - 1 } { q + 1 } $$
Now using induction we show that for every natural number $ n $, we must have $ | k _ q | \ge q ^ n $. For the base case, we note that if $ k _ q = 0 $, then $ k _ { q +1 } $ can't be an integer, so $ | k _ q | \ge 1 = q ^ 0 $. For the induction step, we have:
$$ \frac { | k _ q | + 1 } { q + 1 } \ge \frac { | k _ q - 1 | } { q + 1 } = | k _ { q + 1 } | \ge ( q + 1 ) ^ n $$
$$ \therefore \quad | k _ q | \ge ( q + 1 ) ^ { n + 1 } - 1 \ge q ^ { n + 1 } $$
But this leads to an obvious contradiction. So $ c $ doesn't exist.
A: I found this question very interesting, since it's easy to prove that $c$ is not constant by example meanings, however, the relevant part is to give the proof mathematically, which I think I have found.
We assume that for $q=k$ ; $q=k+1$ and that $c$ is constant, so the following relations should hold:
$0 \equiv \sum_{i=1}^{k}i! - c \pmod k$
$0 \equiv \sum_{i=1}^{k+1}i! - c \pmod{k+1}$
Let's put all together:
$c=\sum_{i=1}^{k}i! -kp$
$0 \equiv \sum_{i=1}^{k+1}i! - \sum_{i=1}^{k}i! -kp \pmod{k+1}$
$0 \equiv (k+1)! + \sum_{i=1}^{k}i! - \sum_{i=1}^{k}i! -kp \pmod{k+1}$
$0 \equiv (k+1)! -kp \pmod{k+1}$
$p=\sum_{i=1}^{k}i! - c$
$0 \equiv (k+1)! -k(\frac{\sum_{i=1}^{k}i! - c}{k}) \pmod{k+1}$
$0 \equiv (k+1)! - \sum_{i=1}^{k}i! - c \pmod{k+1}$ ($*$)
Since $\sum_{i=1}^{k}i! - c$ is multiple of $k$ then for the latter ($*$) to be true we need that:
$GCD(k+1, \sum_{i=1}^{k}i! - c) \neq 1$
Maybe sounds a bit confusing at the first time, but makes sense. This attempt just tell us for $c$ to be constant the sum of the previous factorials minus $c$ has to be multiple of the current modulus.
For example, take k=5
$GCD(k+1, \sum_{i=1}^{k}i! - c) = GCD(6, 153 - 3) = GCD(6,150) = 6$ so both $k$ and $k+1$ will yield same $c$
but for k=6
$GCD(k+1, \sum_{i=1}^{k}i! - c) = GCD(7, 873 - 3) = GCD(7,870) = 1$ thus $c$ in $k$ and $c$ in $k+1$ yield $3$ and $5$ respectively.
Take into account that I have put some effort elaborating this answer, maybe there exists other simple proof, but I least I try to throw some light to the question.
