Power of Ring Negative

Theorem
Let $\left({R,+,\circ}\right)$ be a Definition:Ring.

Let $n \in \N_{>0}$ be a strictly positive integer.

Let $x \in R$.

Then:
 * If $n$ is even, then $\circ^n \left({-x}\right) = \circ^n \left({x}\right)$.
 * If $n$ is odd, then $\circ^n \left({-x}\right) = -\circ^n \left({x}\right)$.

Proof
First, suppose that $n$ is even.

Then for some $m\in \N_{>0}$, $n = 2m = m + m$.

Thus since $\circ$ is associative,
 * $\displaystyle \circ^n \left({x}\right) = \prod_{i=1}^m \left({-x}\right) \circ \left({x}\right)$

By Product of Ring Negatives, $\left({-x}\right) \circ \left({-x}\right) = x \circ x = \circ^2 \left({x}\right)$.

Thus we have
 * $\displaystyle \circ^n \left({-x}\right) = \prod_{i=1}^m \circ^2 \left({x}\right)$

By associativity, then,
 * $\circ^n \left({-x}\right) = \circ^{2m} \left({x}\right) = \circ^n \left({x}\right)$.

Now suppose instead that $n$ is odd.

If $n=1$, then $\circ^n \left({-x}\right) = -x = -\circ^n \left({x}\right)$.

Otherwise,