Sign of Odd Power

Theorem
Let $x \in \R$ be a real number.

Let $n \in \Z$ be an odd integer.

Then:
 * $x^n = 0 \iff x = 0$
 * $x^n > 0 \iff x > 0$
 * $x^n < 0 \iff x < 0$

That is, the sign of an odd power matches the number it is a power of.

Proof
If $n$ is an odd integer, then $n = 2k + 1$ for some $k \in \N$.

Thus $x^n = x \cdot x^{2k}$.

But $x^{2k} \ge 0$ from Even Powers are Positive.

The result follows.