What is the remainder left after dividing $1! + 2! + 3! + ... + 100!$ by $5$? I have this question as a homework. 

What is the remainder left after dividing $1! + 2! + 3! + \cdots + 100!$ by $5$?

I tried this: I noticed that every $n! \equiv 0 \pmod{5}$ for every $n\geq 5$.
For $n < 5$:
$$\begin{align*}
1! &\equiv 1 \pmod{5} \\
2! &\equiv 2 \pmod{5} \\
3! &\equiv 1 \pmod{5} \\
4! &\equiv 4 \pmod{5} \\
\end{align*}$$
So $1! + 2! + \cdots + 100! \equiv 8 \equiv 3 \pmod{5}$.
Therefore the remainder is $3$. 
Am I thinking properly?  
Thanks a lot.

To complement the other answer in less theory-heavy terms, you can view it as:
$$1! + 2! + 3! + 4! + 5! (1 + 6 + [6 \times 7] + \dots + [6 \times \dots \times 100])$$
The final term is clearly divisible by $5$; so we just need to look at $1! + 2! + 3! + 4!$ and its remainder, as you did.
