Is it true that odd number raised to power (natural number) will always be odd? Is it true that odd number raised to power (natural number) will always be odd ? If this is true what is explanation behind this?
Thanks

Your intuition about the binomial theorem was a good one:
$$
(2r - 1)^n = \sum_{k = 0}^n \binom{n}{k} (2r)^k(-1)^{n-k} = (-1)^{n-k} + \sum_{k = 1}^n \binom{n}{k} (2r)^k (-1)^{n-k}.
$$
Every term in the rightmost summation has a factor of two, so that sum is even. The remaining term of $(-1)^{n-k}$ either adds or subtracts $1$, so the overall sum is odd.
