I've seen the arithmetic on the dihedral group $D_{2n}$ written in several different ways, but here's what I'm working with. Say that $r$ is a clockwise rotation by $\frac{2\pi}{n}$ radians and $s$ a reflection through a fixed axis of symmetry. Then the elements of $D_{2n}$ can be written in the form $r^k s^i$ for $k \in \{0,1, \ldots, n-1\}$ and $i \in \{0,1\}$.
Based on this, I was able to prove by induction that for any $k \geq 0$, $r^k s = sr^{-k}$. But I found myself, in computations, wanting to use the fact that $sr^k = r^{-k} s$, but I only proved the above statement for positive $k$. I think I may be overthinking how/whether this follows from the statement I proved. I thought to try induction, but I think I can use "periodicity." That probably is not the correct term, but in $D_{2n}$, I know that $r^k = r^{k+mn}$ for any integer $m$. So $r^{-k} = r^{n-k}$. I think I can say: given $k \geq 0$, I can reduce to the case where $0 \leq k \leq n$ by cancelling extraneous factors, and then invoke the aforementioned statement I proved to say that $r^k s = sr^{-k}$. Then $r^k s = sr^{n-k}$, where $n-k \geq 0$. This gives me a non-negative coefficient on both sides of the equation, which is not what I want.
I'd appreciate some help in pointing out where I've gone wrong.
From $$r^ks=sr^{-k},$$ multiply on the left by $r^{-k}$ and on the right by $r^k$ to get $$\begin{align*} r^ks&=sr^{-k}\\ r^{-k}r^ksr^{k} &= r^{-k}sr^{-k}r^k\\ sr^k&=r^{-k}s \end{align*}$$ which shows how to get from one equality to the other (reverse the steps to show their equivalence). This has the advantage of also holding for the infinite dihedral group, where modular arguments would not.