Simplifying $ 1^5+2^5+3^5+\dots+14^5+15^5 \pmod{13} $ 
$$ 1^5+2^5+3^5+\dots+14^5+15^5 \pmod{13} $$

I found this question on my old textbook, it seems very trivial but my answer and the answer of book are different. My solution is that:
$$ [1^5+2^5+3^5+4^5+5^5+6^5+(-6)^5+(-5)^5+(-4)^5+(-3)^5+(-2)^5+(-1)^5+0^5+1^5+2^5] \pmod{13} $$
So, it gives us $33\pmod{13}=7$, but the answer is $8$. Where am I missing?
Moreover, if you know any trick or shortcut for these types of problem, can you share your knowledge?

Since $5$ is prime to $\phi(13)=12$, each residue $\bmod 13$ is the fifth power of one residue and then
$1^5+2^5+...13^5\equiv1+2+...+13\equiv 78\equiv0\bmod 13$.
So the given sum reduces to $14^5+15^5\equiv1^5+2^5\equiv33\equiv7$.
And $7$ is correct.
Either the book is wrong or, as sometimes happens, you mistakenly read an answer to an adjacent problem.  That, of course, cannot be resolved here.
