It is not as easy as you had expected it to be

Relevant wiki: Euler's Theorem

Let $S = 1^5 + 2^5 + 3^5 + ... + 100^5 = \displaystyle \sum_{n=1}^{100} n^5$ . We note that even $n^5$ is divisible by 4. Therefore, we have:

$\begin{aligned} S & \equiv \left(1^5 + 3^5 + 5^5 + ... + 99^5 \right) \text{ (mod 4)} \end{aligned}$

Since all odd numbers are coprime to 4, we can apply the Euler's theorem. Since the Euler's totient function of 4 is $\phi(4) = 4\left(1-\dfrac 12\right) = 2$ . Then we have:

$\begin{aligned} S & \equiv \left(1^{2\cdot 2+1} + 3^{2\cdot 2+1} + 5^{2\cdot 2+1} + ... + 99^{2\cdot 2+1} \right) \text{ (mod 4)} \\ & \equiv \left(1 + 3 + 5 + ... + 99 \right) \text{ (mod 4)} \\ & \equiv \frac {50(1+99)}2 \text{ (mod 4)} \\ & \equiv \boxed{0} \text{ (mod 4)} \end{aligned}$

It actually is as easy as I thought it to be. Consider fifth powers modulo 4: $0^5 \equiv 0\ \ ;\ \ \ 1^5 \equiv 1\ \ \ ;\ \ 2^5 \equiv 4\cdot 2^3 \equiv 0\ \ \ ;\ \ 3^5 \equiv (-1)^5 \equiv -1.$ The sum modulo 4 is $(1^5 + 2^5 + 3^5 + 4^5) + (5^5 + 6^5 + 7^5 + 8^5) + \cdots \equiv 25\cdot(1 + 0 + (-1) + 0) \equiv 25\cdot 0 = \boxed{0}.$