Computing the order of elements in Dihedral Groups

I am working through "Abstract Algebra" by Dummit & Foote. Exercise 1.2.1 states:

Compute the order of each of the elements in ... $D_6$, $D_8$, and $D_{12}$.

I have found that: $$\begin{align*} D_6 &= \{ 1, r, r^2, s, sr, sr^2 \},\\ D_8 &= \{ 1, r, r^2, r^3, s, sr, sr^2, sr^3 \}, \qquad\text{and}\\ D_{12} &= \{ 1, r, r^2, r^3, r^4, r^5, s, sr, sr^2, sr^3, sr^4, sr^5 \}. \end{align*}$$

In these descriptions, $r$ is rotation of $\frac{2\pi}{n}$ radians, and $s$ is reflection through vertex "1" and the origin.

Also, this book uses $D_{2n}$ for the dihedral group of rigid motions of the regular $n$-gon (that is, the index is the order of the group).

Determining the orders of powers of $r$ is easy (and $|s| = 2$), but I get confused when trying to compute, say, $|sr^5|$.

I think the key is the statement: $r^is = sr^{-i}$ for $0 \leq i \leq n$.

Then we would have $$ (sr^5)(sr^5) = s(r^5s)r^5 = s(sr^{-5})r^5 = (ss)(1) = s^2 = 1 $$

So $|sr^5| = 2$ ?

Am I doing this correctly? Is there a better way?
Thanks.

Yes, you are perfectly right. If you think at the problem geometrically, probably you will get a sharper picture of the situation.

Every element of the form $sr^k$ is in fact a reflection around some axes, thus it's very natural to expect it has order 2. Your proof is correct, indeed; just replace the $5$ with a $k$ and you've done!

bye