To prove that since $1+2+\cdots+n=\frac{n(n+1)}{2}$, the problem is equivalent to proving: $n(n+1)$ divides $2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$. Because $n$ and $n+1$ are coprime, this is equivalent to proving
$$n \mid 2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$$

and
$$(n+1) \mid 2\left(1^{k}+2^{k}+\cdots+n^{k}\right)$$

In fact, since $k$ is odd, by the factorization formula (6) from the first unit, we know that
$$\begin{aligned} & 2\left(1^{k}+2^{k}+\cdots+n^{k}\right) \\ = & {\left[1^{k}+(n-1)^{k}\right]+\left[2^{k}+(n-2)^{k}\right]+\cdots+\left[(n-1)^{k}+1^{k}\right]+2 n^{k} } \end{aligned}$$

is a multiple of $n$. Similarly,
$$2\left(1^{k}+2^{k}+\cdots+n^{k}\right)=\left[1^{k}+n^{k}\right]+\left[2^{k}+(n-1)^{k}\right]+\cdots+\left[n^{k}+1^{k}\right]$$

is a multiple of $n+1$.