I did some calculations and found out that for $n=1,2,3,4$ the remainders are $35,1,31,19$ respectively. From then on I conjectured that the remainder was always 7. I then attempted to prove this by simple induction on $n$. For $n=5$ it checked out.
$$\sum_{i=1}^{n+1} (-1)^i i! = \bigg( \sum_{i=1}^{n} (-1)^i i! \bigg) + (-1)^{n+1}(n+1)!$$
But this amounts to proving that $36\mid (-1)^{n+1}(n+1)!$ for all $n\geq 5$. We can do simple induction on $n$ a second time. This new hypothesis is true for 5. And
$$(-1)^{n+2}(n+2)! = (-1)(-1)^{n+1}(n+2)(n+1)! = k(-1)^{n+1}(n+1)!$$
Which follows from the induction hypothesis and the properties of divisibility. Hence the remainder is 7 for all $n\geq 5$.
Is this reasoning more or less correct?