Let $m$ and $n$ be odd integers. Prove $ \sum_{i=1}^{m}i^n≡ 0 \pmod{m}$ Let $m$ and $n$ be odd integers. Prove $ \sum_{i=1}^{m}i^n≡ 0 \pmod{m}$ 
Here's my attempt at an Induction Proof:
Let this
$$ \sum_{i=1}^{m}i^n≡ 0  \pmod{m} $$ 
be correct for some $m$. Now let's prove that's correct also for $m+1$:
$$ \sum_{i=1}^{m+1}i^{n} = i^1+\dots+i^m+i^{m+1} ≡ 0  \pmod{m} $$ 
and then this $i^1+\dots+i^m$ is divisible by $m$ because the first part of induction.
What next? Can anyone help me?
 A: Let $m,n$ be odd integers, then we want to show $$\sum_{i=1}^{m}i^n \equiv 0  \pmod{m}.$$

Lets look at an example $$1^7 + 2^7 + 3^7 + 4^7 + 5^7 \equiv 0 \pmod 5.$$
Clearly we can throw away the last term: $$1^7 + 2^7 + 3^7 + 4^7 \equiv 0 \pmod 5$$
Now notice that half the terms may be written as negatives $$1^7 + 2^7 + (-1 \cdot 2)^7 + (-1)^7 \equiv 0 \pmod 5$$
and we can pull that sign out when we have an odd exponent $$1^7 + 2^7 - (2^7 + 1^7) \equiv 0 \pmod 5$$ which is obviously true.

this inspires a proof: $$\sum_{i=1}^{m}i^n \equiv \sum_{i=1}^{m-1} i^n \equiv \left(\sum_{i=1}^{(m-1)/2} i^n\right) + \left(\sum_{i=(m+1)/2}^{m-1} i^n\right)$$ $$\equiv \left(\sum_{i=1}^{(m-1)/2} i^n\right) + \left(\sum_{i=1}^{(m-1)/2} (m-i)^n\right)\equiv \left(\sum_{i=1}^{(m-1)/2} i^n\right) - \left(\sum_{i=1}^{(m-1)/2} i^n\right)\equiv 0 .$$
A: $\begin{array}\\
i^n+(m-i)^n
&=i^n+\sum_{j=0}^n \binom{n}{j}m^j(-1)^{n-j}i^{n-j}\\
&=i^n+(-1)^ni^n+\sum_{j=1}^n \binom{n}{j}m^j(-1)^{n-j}i^{n-j}\\
&=\sum_{j=1}^n \binom{n}{j}m^j(-1)^{n-j}i^{n-j}
\qquad\text{since }n\text{ is odd}\\
&=m\sum_{j=1}^n \binom{n}{j}m^{j-1}(-1)^{n-j}i^{n-j}\\
&≡ 0  \pmod{m}\\
\end{array}
$
Since
$\sum_{i=1}^{m-1}i^n
=\sum_{i=1}^{m-1}(m-i)^n
$,
$2\sum_{i=1}^{m-1}i^n
≡ 0  \pmod{m}
$.
Since $m$ is odd,
this implies that
$\sum_{i=1}^{m-1}i^n
≡ 0  \pmod{m}
$.
(I know you can skip this
by pairing symmetric terms,
but,
what the heck.)
Note that the term
$m^n$ can be ignored.
