This year is odd

Relevant wiki: Euler's Theorem

We find that $2017$ is a prime number. Thus we can say that $\text{gcd } (a,2017) = 1$ where $a < 2017$ is a positive integer. Thus, we can apply Euler's theorem as

$a^{2017} \pmod{2017} \equiv a^{2017 \pmod{\phi (2017)}} \pmod{2017} \equiv a^{2017 \pmod{2016}} \pmod{2017} \equiv a \pmod{2017}$

Thus

$\begin{aligned} \displaystyle \sum_{a=1}^{2016} a^{2017} \pmod{2017} &\equiv \sum_{a=1}^{2016} a \pmod{2017} \\ & = 1008 \times 2017 \pmod{2017} \\ & \equiv \boxed{0} \pmod{2017} \end{aligned}$