Power of Ring Negative

Theorem
Let $\struct {R, +, \circ}$ be a ring.

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

Let $x \in R$.

Then:
 * If $n$ is even:
 * $\map {\circ^n} {-x} = \map {\circ^n} x$


 * If $n$ is odd:
 * $\map {\circ^n} {-x} = -\map {\circ^n} x$

Proof
First, suppose that $n$ is even.

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

Thus since $\circ$ is associative:
 * $\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \paren {-x} \circ \paren {-x}$

By Product of Ring Negatives:
 * $\paren {-x} \circ \paren {-x} = x \circ x = \map {\circ^2} x$

Thus:
 * $\ds \map {\circ^n} {-x} = \prod_{i \mathop = 1}^m \map {\circ^2} x$

By associativity:
 * $\map {\circ^n} {-x} = \map {\circ^{2 m} } x = \map {\circ^n} x$

Now suppose instead that $n$ is odd.

If $n = 1$, then:
 * $\map {\circ^n} {-x} = -x = -\map {\circ^n} x$

Otherwise: