Note that $5 + 5^2 + 5^3$ is divisible by $31$

5+5^2+5^3 = 0 mod 31 and 5^3 = 1 mod 31 so done Sniped

Note that $S=\tfrac{5^{2017}-5}{4}$ and $5^{30} \equiv 1 \pmod{31}$. So $5^{2017}-5 \equiv 5^{7}-5 \pmod{31}$. It's easy to verify that this expression is divisible by 31.

Notice that: $5+5^2+5^3 \equiv 0 \pmod{31}$ $5^3 \equiv 1 \pmod{31}$ Therefore $31|5^{3k-2}+5^{3k-1}+5^{3k}$, where k is a natural number Since $3|2016$, we have: $(5+5^2+5^3)+...+(5^{2014}+5^{2015}+5^{2016}) \equiv 0 \pmod{31}$ Therefore, $31|5+5^2+5^3+...+5^{2016}$ $\blacksquare$

From the expansion $a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\cdots+ab^{n-2}+b^{n-1})$ for $a=5^3$, $b=1$ and $n=672$ we get that $$5^3-1 | 5^{2016}-1$$and so $$31 | \frac{5^{2016}-1}{4}$$and finally $31 | 5+5^2+\cdots+5^{2016}$, which is indeed what we wanted.