Proof: Raising a complex number to a rational power

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The problem from the textbook is:

Prove that if (a complex number) $z$ is a number on the unit circle, then $z$ has finitely many distinct powers $z^n$ if and only if the argument of $z$ is a rational multiple of $2 \pi$.

So I understand that $z$ raised to a rational argument $m/n$ can be viewed as trying to find the $n$ roots of $z^m$, which will give me $n$ answers. I also understand that the $n$ numbers $$ e^{i(\frac{\theta}{n}+ \frac{2k\pi}{n})} $$ are evenly spaced around the unit circle, with each successive pair separated by an angle of $$\frac{2\pi}{n}.$$

I'm basically confused about where the "if and only if" statement comes into play. Am I on the right track so far?

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We can write in the form $\;z=e^{it}\;,\;\;t\in\Bbb R\;$ , so

$$z^n=1\iff e^{int}=1\iff nt=2k\pi\;\;\;k\in\Bbb Z\iff t=\frac kn2\pi$$