Sign of Odd Power

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

Let $$n \in \mathbb{Z}$$ 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 $$k \in \mathbb{N}$$.

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

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

The result follows.