$\sum z_i^k=0$ for all $k\ge 2$ implies $z_1=\dots=z_n=0$? 
Let $z_1, z_2,\dots,z_n \in \mathbb C$ such that 
  $$z_1^k+\dots+z_n^k=0$$ for all integers $k\geq 2$.
  Then how to prove that $z_1=z_2=\dots=z_n=0$?

My try:
I got to work it for $n=2$ case, by using some brute force calulation. 
i.e. taking $z_j= r_j e^{i\theta_j}$ and solving further.
But for larger $n$ I don’t have any idea.
It seems like induction argument will work. Any hint?
 A: With a slight generalization of copper.hat's solution to


*

*Does $\lambda_1^n+ \lambda_2^n+ \dots +\lambda_k^n =0 $ for all $n$ imply that $\lambda_1= \lambda_2= \dots= \lambda_k = 0 $?
one can proceed as follows:
Let $p(z) = \sum_{k=0}^N a_k z^k$ be a polynomial such that
$p(0) = 0$, $ p'(0) = 0$, and
$$
p(z_j) = |z_j| \quad \text{for } j = 1, \ldots, n \, .
$$
Then $a_0=a_1=0$ and
$$
 \sum_{j=1}^n |z_j| = \sum_{j=1}^n p(z_j) = \sum_{k=2}^N a_k \sum_{j=1}^n z_j^k = 0
$$
and therefore $z_1 = \ldots = z_n = 0$.
Remark 1: The existence of such a polynomial $p$ can be shown with a slightly 
modified Langrange interpolation: 
Let $w_1, \ldots, w_m$ be the non-zero distinct values in $\{ z_1, \ldots, z_n \}$,
and
$$
 L_j(z) = \prod_{\substack{l=1 \\ l \ne j}}^m \frac{z-w_l}{w_j-w_l}
 \quad (1 \le j \le m)
$$
the corresponding Language polynomials. Then
$$
 p(z) = \sum_{j=1}^m |w_j|\frac{z^2}{w_j^2} L_j(z) \, .
$$
has the desired properties.
Remark 2: The degree of $p$ is at most $m + 1 \le n+1$.
Therefore it suffices to require that
$ z_1^k+ \ldots +z_n^k=0$ for $2 \le k \le n+1$.
A: Here is a proof if we assume vanishing for any consecutive sequence of $n$ exponents, $l\leq k <l+n$.
Let $w_j$,  $j=1, \ldots , t$ be the distinct values amongst $z_1, \ldots , z_n$
and let $w_j$ occur $m_j>0$ many times. 
Then we have the matrix equation
$$\begin{pmatrix}
w_1^l&w_2^l&\cdots &w_t^l\\
w_1^{l+2}&w_2^{l+2}&\cdots &w_t^{l+2}\\
&&\vdots &\\
w_1^{l+t}&w_2^{l+t}&\cdots &w_t^{l+t}\\
\end{pmatrix}\begin{pmatrix}
m_1\\
m_2\\
\vdots \\
m_t\\
\end{pmatrix}=\begin{pmatrix}
0\\
0\\
\vdots\\ 
0\\
\end{pmatrix}$$
which implies that 
$$\det \begin{pmatrix}
w_1^l&w_2^l&\cdots &w_t^l\\
w_1^{l+2}&w_2^{l+2}&\cdots &w_t^{l+2}\\
&&\vdots &\\
w_1^{l+t}&w_2^{l+t}&\cdots &w_t^{l+t}\\
\end{pmatrix}=\prod\limits_j w_j^l\prod\limits_{i<j}  (w_j-w_i)=0$$
Since the $w_j$ are distinct, there is at least one $j$ such that $w_j=0$. By induction the result follows. 
Note that consecutivity of the exponents $k$ is essential since if $\zeta_p$ is a primitive $p$ th root of unity then 
$$\sum\limits_{i=0}^{p-1} (\zeta^i)^{pn+l}=0$$ for all $l$ not a multiple of $p$.   
