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 the mapping defined by
 * $f(x) = \circ^n (x)$.

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

Proof
By Power Function Strictly Increasing on Positive Elements, if $0 < x < y$ then $f \left({x}\right) < f \left({y}\right)$.

Suppose that $x < y < 0$.

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

By Power Function Strictly Increasing on Positive Elements (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, $f_n\left({x}\right) < f_n\left({y}\right)$.

By Sign of Odd Power,
 * $f_n \left({x}\right) < 0 = f_n \left({0}\right)$ when $x < 0$ and
 * $f_n \left({0}\right) = 0 < f_n \left({x}\right)$ when $0 < x$.

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