Power of Ring Negative

Theorem
Let $\left({R, +, \circ}\right)$ be a 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:
 * $\displaystyle \circ^n \left({x}\right) = \prod_{i \mathop = 1}^m \left({-x}\right) \circ \left({x}\right)$

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

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

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

Now suppose instead that $n$ is odd.

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

Otherwise: