TITLE: Showing that $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$
QUESTION [22 upvotes]: For any odd positive integer $k\geq1$, the sum $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$.

I used induction principle for the solution but cannot prove it.
I took $P(k) = 1^k+2^k+\dots+n^k$.
For $P(1)$ it is true.
For $P(n)$ let it be true.
But for $P(n+1)$ I cannot solve it.

REPLY [2 votes]: Since $1+2+...+n = \dfrac{n(n+1)}{2}$, you can prove separately that $S_k = 1^k+2^k+...+n^k$ is divisible by both $n$ and $\dfrac{n+1}{2}$ for $n$ odd, $n/2$ and $n+1$ for $n$ even. 
For instance, when $n$ is odd, prove that $n+1|a^k+(n-a+1)^k$ and then partition $S_k$ in such a way that every summand is divisible by $n+1.$ The rest is pretty much similar for the different cases. 
You can also use a more powerful apparatus called Bernoulli numbers and how they are related to power sums. In fact, you can prove that $S_k$ is a polynomial of $S_1$ with rational coefficients.