How to prove sum of powers property of roots of unity? We know that $1+\alpha_1+\alpha_2+...+\alpha_{n-1}=0$ where $\alpha_i$ are the roots of $z^n=1$. How can I prove that:
$$1+\sum_{i=1}^{n-1}\alpha_i^m=\begin{cases}0\quad m\in Z,m\not\equiv0\pmod n\\n\quad m\in Z,m\equiv0\pmod n\\\end{cases}$$
 A: The roots of $z^n=1$ are $\alpha_k= \omega^k$, where $\omega = \exp(2\pi i/n)$.
When $m$ and $n$ are coprime, the map $z \mapsto z^m$ permutes these roots and so
$$
1^m+\alpha_1^m+\alpha_2^m+\cdots+\alpha_{n-1}^m
=1+\alpha_1+\alpha_2+\cdots+\alpha_{n-1}
=0
$$
When $m$ is a multiple of $n$, the map $z \mapsto z^m$ is the constant map $1$ on these roots and so
$$
1^m+\alpha_1^m+\alpha_2^m+\cdots+\alpha_{n-1}^m
=1+1+1+\cdots+1
=n
$$
When $1 < \gcd(m,n)=d< n$, you get $d$ sums of the same form, but now for $n/d$-th roots of unity and so it's $0$ again, by the first case.
For instance, take $n=6$ and $m=2$. Then
$$
\omega^0+\omega^2+\omega^4+\omega^6+\omega^8+\omega^{10}
=
\omega^0+\omega^2+\omega^4+\omega^0+\omega^2+\omega^{4}
=
2(\lambda^0+\lambda^1+\lambda^2)
$$
where $\lambda=\omega^2$ is the primitive cubic root of unity.
A: Note we can write $\alpha_k=e^{i\frac{2\pi}{n}k}$
then if $m$ is a multiple of $n$ we have $\alpha_k^m=1$ thus then we have $$1+\sum_{k=1}^{n-1} 1=n$$
otherwise we get that $\alpha_1^m \neq 1$. Then we get 
\begin{align*}
1+ \sum_{k=1}^{n-1} \alpha_k^m
&=  \sum_{k=0}^{n-1} e^{i\frac{2\pi}{n}mk}\\
&=\frac{1-e^{i2\pi m}}{1-e^{i\frac{2\pi}{n}m}}\\
&= \frac{0}{1-e^{i\frac{2\pi}{n}m}}=0
\end{align*}
