Showing that $1^k+2^k + \dots + n^k$ is divisible by $n(n+1)\over 2$ 
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.
 A: Let $\Pi_n$ be the vector space of polynomials with real coefficients and degree $\le n$.
Consider the map $\Delta: \Pi_{k+1} \to \Pi_{k}$ given by $(\Delta P)(x) = P(x+1)-P(x)$. This is a linear map from an $k+2$-dimensional vector space to a $k+1$-dimensional vector space with a $1$-dimensional kernel consisting exactly of the constant functions. Hence, by the rank-nullity-theorem $\Delta$ is surjective. Thus there is a $Q \in \Pi_{k+1}$ such that $Q(x+1)-Q(x)=(x+1)^k$. Define $P(x)=Q(x)-Q(0)$. Then for each $n \in \Bbb N$:
$$P(n)=Q(n)-Q(0)=n^k+Q(n-1)-Q(0)=n^k+(n-1)^k+Q(n-1)+Q(0)=\ldots=n^k+(n-1)^k+ \ldots +1^k+Q(0)-Q(0)=n^k+(n-1)^k+\ldots+1^k$$.
Since $P(0)=P(-1)=0$, $X$ and $X+1$ divide $P(X)$, $X(X+1)|P(X)$. Hence $P(n)/(n(n+1))$ is a rational whose denominator can be assumed to have a fixed value $q$. On the other hand, choosing $n$ so large that $n(n+1)$ has only the divisor 2 smaller than $q$,you see that the denominator of $P(n)/(n(n+1))$  must be $2$ hence it must always be $2$ or $1$ and $n(n+1)/2|P(n)$.
A: As $k$ is odd, 
$r^k+(n+1-r)^k$ is divisible by $r+(n+1-r)=n+1$
and $s^k+(n-s)^k$ is divisible by $s+(n-s)=n$
Putting $r=1,2,\cdots,n-1,n$ and adding them we get
$$(n+1)|\sum_{1\le r\le n}\{r^k+(n+1-r)^k\}  \implies (n+1)|2\sum_{1\le r\le n}r^k$$
Putting $s=1,2,\cdots,n-1,n$ and adding them we get
$$n|\sum_{1\le s\le n}\{s^k+(n-s)^k\}  \implies (n)|2\sum_{1\le s\le n}s^k \implies n|2\sum_{1\le r\le n}r^k$$
Now, either $n$ or $n+1$ is odd
If $n$ is odd, $$n|2\sum_{1\le r\le n}r^k\implies n|\sum_{1\le r\le n}r^k$$
$$\implies \text{lcm}(n+1,n) | 2\sum_{1\le r\le n}r^k$$
$$\implies n(n+1) | 2\sum_{1\le r\le n}r^k\implies \frac{n(n+1)}2 | \sum_{1\le r\le n}r^k$$
Similarly when $n+1$ is odd
A: For odd $k$ we have that
$$
a^k+b^k=(a+b)(a^{k-1}-a^{k-2}b+a^{k-3}b^2-\dots+b^{k-1})
$$
Thus, each column of
$$
\begin{align}
&0^k+\hphantom{(n-\ )}1^k+\hphantom{(n-\,\,)}2^k+\dots+n^k\\
&n^k+(n-1)^k+(n-2)^k+\cdots+0^k
\end{align}
$$
is divisible by $n$ and each column of
$$
\begin{align}
&1^k+\hphantom{(n-\ )}2^k+\hphantom{(n-\,\,)}3^k+\dots+n^k\\
&n^k+(n-1)^k+(n-2)^k+\cdots+1^k
\end{align}
$$
is divisible by $n+1$. Since $(n,n+1)=1$ we get that
$$
n(n+1)\,|\,2(1^k+2^k+3^k+\dots+n^k)
$$
and therefore, since $n(n+1)$ is even,
$$
\left.\frac{n(n+1)}{2}\middle|1^k+2^k+3^k+\dots+n^k\right.
$$
A: 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. 
