Odd Power Function is Strictly Increasing/General Result

Theorem
Let $\left({R,+,\circ,\le}\right)$ be a totally ordered ring.

Let $n$ be a odd positive integer.

Let $f \colon R \to R$ be defined by
 * $f(x) = \circ^n (x)$.

Then $f$ is strictly increasing on $R$.

Proof
Trivial case: if $n = 1$, then $f$ is the identity mapping, and Identity Mapping is Order Isomorphism. For the remaining portion of the proof, suppose that $n > 1$.

Lemma
Let $0 < x < y$ and $0 < z < w$

By Properties of Ordered Ring$(6)$,
 * $z \circ x < z \circ y$
 * $z \circ y < w \circ y$

Then $z \circ x < w \circ y$ by transitivity.

Lemma 2
If $0 < x < y$, then $0 < \circ^n \left({x}\right) < \circ^n \left({y}\right)$.

Proof
Apply lemma 1 inductively.

Thus $f$ is strictly increasing on $\left({0 \,..\, \to}\right)$.

Now suppose that $x < y < 0$.

By Properties of Ordered Ring, $0 < -y < -x$.

By the case for strictly positive $x$ and $y$ (applied to $-y$ and $-x$),


 * $0 < f_n \left({-y}\right) < f_n \left({-x}\right)$.

By Power of Ring Negative, $f_n \left({-x}\right) = -\left({x}\right)$ and $f_n \left({-y}\right) = -f_n\left({y}\right)$.

Thus $0 < -f_n \left({y}\right) < -f_n\left({x}\right)$.

By Properties of Ordered Ring, $f_n\left({x}\right) < f_n\left({y}\right) < 0$.

Thus we have shown that $f_n$ is strictly increasing on the positive elements and the negative elements, and across zero.