I wish to prove that for $n,k\in\mathbb{N} > 1$, we can always write $n^k$ as a sum of $n$ odd positive integers.
I have an idea of how to approach this, but my method seems to cumbersome. I am tempted to say that this sum can be achieved by peeling the $k$-dimensional cube of side length $n$, for example: $n^2 = \sum_{i=1}^{n}2i-1$ by taking the layers of the $n\times n$ square and $n^3 = \sum_{i=1}^{n}3i^3 - 3i + 1$ by taking the layers of the $n\times n\times n$ cube, but that seems overkill and it is tedious to do in the general case.
Does anyone have a more elegant proof for this statement?