Split n^m into n numbers, the average of the numbers would be n^{m-1}.
If n is odd, we can produce the consecutive odd numbers {…n^{m-1}-4, n^{m-1}-2, n^{m-1}, n^{m-1}+2, n^{m-1}+4…}. For example, for 3^3, it will be {7, 9, 11}.
If n is even, we can produce the consecutive odd numbers {…n^{m-1}-5, n^{m-1}-3, n^{m-1}-1, n^{m-1}+1, n^{m-1}+3, n^{m-1}+5…}. For example, for 4^3, it will be {13, 15, 17, 19}.
It can be written as:
n^m = \sum_{k=1}^n {N^{m-1}}+2k-n-1
When n is odd, N^{m-1} is odd, 2k-n-1 is even, the sum is odd.
When n is even, N^{m-1} is even, 2k-n-1 is odd, the sum is odd.
Therefore, we have consecutive odd numbers for both scenarios.