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If $p$ is a prime number, how can you show that there are exactly $p^{n-1}(p-1)$ primitive $p^n$-th roots of unity?

I am a little stuck on how to begin this proof. Do you need to use orders or consider roots of unity in polar terms? Any help would be greatly appreciated. Thanks!

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Note that $\varphi(p^n)=(p-1)p^{n-1}$. Since $$\zeta_k=\exp\frac{2\pi k}{p^n}$$ is an $p^n$-th root of unity iff $(k,p^n)=1$, you get your result.

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I am not exactly sure what the group contains. Could you elaborate? – RXY15 Feb 21 '14 at 3:31
@RXY15 I initially misread your question. – Pedro Tamaroff Feb 21 '14 at 3:31
@RXY15 Euler's totient. – Pedro Tamaroff Feb 21 '14 at 3:33
Hmm I haven't learned that yet but I will look into it. Thanks! – RXY15 Feb 21 '14 at 3:33

There are $p^n$ $p^n$th roots of unity. The ones that aren't primitive are the ones that are also $p^{n-1}$th roots of unity, and there are $p^{n-1}$ of them. So, the number of primitive ones in $p^n-p^{n-1}$.

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