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 $n &lt; 5$:

$$\begin{align*}
1! &amp;\equiv 1 \pmod{5} \\
2! &amp;\equiv 2 \pmod{5} \\
3! &amp;\equiv 1 \pmod{5} \\
4! &amp;\equiv 4 \pmod{5} \\
\end{align*}$$

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.
You are right. 

Notice that the unit digit of the sum is $3$, because all the terms in the sum end up with a zero from $5!$ and hence to find the unit digit, we just need to find the unit digit of the sum $1!+2!+3!+4!=33$ which is $3$.

So, when the sum $1! + 2! + 3! + \cdots + 100!$ is divided by $5$, it leaves a remainder $3$.