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
 A: 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.
A: Let $a^0=1$ be the base case. Then apply the inductive step $f(x)=ax$ (observe that $f^n(a)=a^n$). Clearly, this map preserves the parity, thus it must be odd.
A: As some users are mentioning in the comments, a simpler (and, I think, more enlightening) way to approach this problem would be by induction. In fact, I'm going to use a slightly sloppy form of induction that I hope decreases the formality (in case you aren't familiar with induction) and increases the clarity.
Let's start with any two odd numbers: $2j - 1$ and $2k - 1$ for natural $j$ and $k$. Their product is
$$
(2j - 1)(2k - 1) = 4jk - 2j - 2k + 1 = 2(2jk - j - k) + 1.
$$
Since I've written the product as twice a natural number plus one, the product must be odd.
Now to your question about $(2j - 1)^n$ for natural $n$. Write this as
$$
\underbrace{(2j - 1) \cdots (2j -1)}_{n \text{ copies}}.
$$
From our previous observation, $(2j - 1)(2j - 1)$ is odd, so we have
$$
(\text{some odd number}) \cdot \underbrace{(2j - 1) \cdots (2j -1)}_{n-2 \text{ copies}}.
$$
We can repeat since the product of any two odds is odd. Swallowing up the next $2j-1$, we get
$$
(\text{some other odd number}) \cdot \underbrace{(2j - 1) \cdots (2j -1)}_{n-3 \text{ copies}}
$$
and so on until we're left with one giant odd number in the end.
