Because $2017$ is odd, we always have \begin{align} 1^{2017} + (n+1)^{2017} &= (n+2)(\cdots)\\ 2^{2017} + n^{2017} &= (n+2)(\cdots)\\ 3^{2017} + (n-1)^{2017} &= (n+2)(\cdots)\\ \cdots\\ (n+1)^{2017} + 1^{2017}&=(n+2)(\cdots) \end{align} where $(\cdots)$ are some polynomials of $n$. Adding these $(n+1)$ equations together yields $$2\cdot\Big(1^{2017}+2^{2017}+\cdots+n^{2017}+(n+1)^{2017}\Big)=(n+2)(\cdots)$$ If is easy to show that $(\cdots)$ above is always even regardless of $n$'s parity. Hence, $\Big(1^{2017}+2^{2017}+\cdots+n^{2017}+(n+1)^{2017}\Big)$ must a multiple of $(n+2)$. However because $(n+1)$ and $(n+2)$ are relatively prime to each other, $(n+1)^{2017}$ is not a multiple of $(n+2)$. Therefore, $\Big(1^{2017}+2^{2017}+\cdots + n^{2017}\Big)$ cannot be a multiple of $(n+2)$.