Prove n-th root is different each other. 
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*I already know FTA and I have a question about solving $z^n = re^{ix}$. The candidates for the roots of this equation are $z=r^{\frac{1}{n}}e^{i\frac{2k\pi+x}{n}}$. The number of roots should be below $n$ by FTA, but to verify these candidates are all roots and nothing else can't be, the $z$'s which are depending on $k$ are different each other. I don't know how to prove $z$'s are distinct.

*Let $n$ is odd and $\zeta_n = e^{i\frac{2\pi}{n}}$. Surely, $\zeta_n^n$. I wonder $\zeta_n$ cannot be 1 before the power of $n$. That is $\zeta_n^2, \ldots \zeta_n^{n-1}$ cannot be 1. Is it right?. How can we prove it?
 A: Hints:


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*$\;z^n=a \iff p(z)=z^n - a = 0\,$, and if $\,p(z)$ had a multiple root, then that root would have to be a common root with its derivative $p'(x)=n\,z^{n-1}\,$.

* Consider for example the case $\,n=4\,$ with $\,z^4=1\,$, where $\pm 1$ are also roots of $z^2=1$.

[ EDIT ]  I misread OP's point 2. as "if $\zeta_n$ is any $n^{th}$ root of unity, then is it possible that $\zeta_n^k = 1$ for some power $1 \le k \le n-1$ ?", and the second hint (now crossed out) pointed out that the answer is "yes" in general.

However, the question actually being asked was whether this can happen for the particular root $\zeta_n = e^{i\frac{2\pi}{n}}$. The answer to that is "no", which follows from the more general statement that the primitive roots of unity are those $e^{i \frac{2k\pi}{n}}$ with $\gcd(k,n)=1\,$. Since $\zeta_n$ corresponds to $k=1$ and $\gcd(1,n)=1\,$, $\zeta_n$ is always a primitive root of unity i.e. the lowest power with ${\zeta_n}^m = 1$ is $m=n$.
A: You can actually derive (1) from (2), if (2) is proven for all n.
Namely, if $z_k=r^{\frac {1}{n}}e^{i\frac {2k\pi+x}{n}}$ then $\frac {z_k}{z_l}=e^{(k-l)\frac {2i\pi}{n}}=\zeta_n^{k-l} $ for $0\le l\le k\lt n $. Thus, it suffices to show that $\zeta_n^k=1$ (where $0\le k\lt n $) only for $k=0$.
To prove (2), note that $\zeta_n^k=\cos\frac {2k\pi}{n}+i\sin\frac {2k\pi}{n} $. The real part must be $1$, and $ \cos\alpha=1$ only for $\alpha=2m\pi $ for some  integer $m $, so it follows that $\frac {k}{n} $ is an integer. This is possible only for $k=0$.
