Inductive Proof regarding symmetries of the square

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I am working through some exercises in the free online book "Algebra: Abstract and Concrete - Edition 2.6" by Frederick M. Goodman, a link to which can be found here Algebra: Abstract and Concrete in preparation for a course in Abstract Algebra come this fall. I am working through exercise 1.3.2 (part a) which asks:

Consider the symmetries of the square card: Show that any positive power of $r$ must be one of {$e, r, r^2, r^3$}. First work out some examples, say through $r^{10}$. Show that for a natural number $k$, $r^k = r^m$ where $m$ is the nonnegative remainder after division of $k$ by 4.

So for the first part I simply wrote out the powers of $r$ $\geq 4$ using the following logic:

  • $r^4 = e$ since $r^4 = 4*(\pi/2) = 2\pi$ based on the fact that $r$ is defined as a rotation of $\pi/2$ about the axis through the centroid of the faces of the square.
  • $r^5 = r*r^4 = r*e = r$
  • $r^6 = r*r^5 = r*r = r^2$
  • And so on up to the tenth power

For the actual problem statement I thought it would be appropriate to use proof by induction, considering the question is asking us to prove a statement about the natural numbers.

  • I took the base case to be $k = 4$ since $4$ is the first non-unique power of $r$ and showed that $r^4 = r^0 = e$ since $4/4 = 1 + 0$
  • The inductive hypothesis would then be: Assume $r^n = r^m$ for $4 \leq n \leq k$ where m is the nonnegative remainder after division of k by 4.
  • Then we would want to show that $r^{k+1} = r^{m+1}$ correct? For that step I just did: $r^{k+1} = r^k * r = r^m * r$ (by inductive hypothesis) = $r^{m+1}$

Would that be correct? I haven't done many proofs by induction, at least in quite some time, and I'm feeling like that was a little bit too simplistic.