<s>[INST] Proof that $n+1$ divides $\sum_{k=1}^n k^q$ for $q > 0$ Since $$\sum_{k=1}^n k = \frac{n\left(n+1\right)}{2}$$
$$\sum_{k=1}^n k^2 = \frac{n\left(n+1\right)\left(2n+1\right)}{6}$$
$$\sum_{k=1}^n k^3 = \frac{n^2\left(n+1\right)^2}{4}$$
... and so on. Is there a proof that generally $\left(n+1\right) | \sum_{k=1}^n k^q$ for $q\in\mathbb{N}$ positive?   
 [/INST] I think what you claim in your comments is true. We can use induction. Let's assume that all sums up to $m$ contains this term $(n+1)$.
\begin{align*}
(n+1)^{m+2} - 1&= 
\sum_{k=1}^n (k+1)^{m+2} - \sum_{k=1}^n k^{m+2}  \\ 
 &= \sum_{k=1}^n \left( 1 + k + k^2 + \dots + k^{m+1} \right )\\ 
 &= n + (n+1) \left( \text{ some junk term } \right )+\sum_{k=1}^n k^{m+1}\\ 
\end{align*}
which gives 
$$(n+1)^{m+2} - (n+1) -(n+1)  \left( \text{ some junk term } \right ) = \sum_{k=1}^n k^{m+1} $$
</s>