This is a problem from Terence Tao's Solving mathematical problems, a personal perspective. The problem is:
Let $k,n\in\mathbb{N}$ with $k$ odd. Prove that the sum $1^k+2^k+\cdots+n^k$ is divisible by $1+2+\cdots+n$.
Here is the text's solution. However, I thought that the second modulus congruence that goes on like:
$1^k+2^k+3^k+\cdots+(m-1)^k+0^k +1^k+\cdots+(m-1)^k + 0\pmod m$."
Could be simplified as
$1^k+2^k+3^k+\cdots+(m-1)^k+0^k+(-(m-1))^k+\cdots+(-1)^k+0 \pmod m$ Now since $k$ is odd, $(-1)^k$ has to be $-1$. Similarly, $(-(m-1))^k$ has to be $-(m-1)^k$. Therefore, the first $m-1$ terms cancel with the last $m-1$ terms.