Sum of powers of complex numbers equals zero implies the numbers themselves are zero

Question

Suppose we have the complex numbers $x_1,x_2,\dots,x_N$ and that we know that
$$x_1+\dots+x_N =0,$$
$$x_1^2+\dots+x_N^2 =0,$$
$$\dots$$
$$x_1^N+\dots+x_N^N =0.$$
Does this imply that all $x_m=0$? There should be enough equations to get a unique solution. I have proven it in the case $N=2$, but going further requires a lot of algebra, and I'm wondering if it can be seen in an easier way than solving the system of equations by brute force.

Answer 1

Yes. By the Newton-Girard formulas the fact that $p_1=p_2=\ldots=p_N=0$ implies the fact that $e_1=e_2=\ldots=e_N=0$, so the variables $x_1,\ldots,x_N$ are roots of the polynomial $z^N$, i.e. they are all zero.

A celebrated consequence is the following: if $M$ is a $N\times N$ real symmetric matrix such that $\text{Tr}(M)=\text{Tr}(M^2)=\ldots=\text{Tr}(M^N)=0$, then $M=0$.