Determine the last digit of $1! + 2! + 3! + 4! + ⋯ + 1000!$.
We first notice that, for some integer $k$,
$$1! + 2! + 3! + 4! + ⋯ + 1000! = 1! + 2! + 3! + 4! +10k $$
since all the terms including and after 5! include factors 5 and 2, and so are multiples of 10.
Also
$$ 10k \equiv 0 \pmod {10} $$
This means
$$ \begin{align} 1! + 2! + 3! + 4! + ⋯ + 1000! & = 1! + 2! + 3! + 4! +10k \\ \\ & \equiv 1! + 2! + 3! + 4! + (0) \pmod{10} \\ \\ & \equiv 33 \pmod{10} \\ \\ & \equiv 3 \pmod{10} \end{align}$$
So the last digit of $1! + 2! + 3! + 4! + ⋯ + 1000!$ is 3.