System of power sums in complex numbers

Given $n$ complex numbers $z_1,z_2,\ldots,z_n$ such that $$z_1^k+z_2^k+\cdots +z_n^k=0$$ for all $k\in\mathbb{N}$. Prove that $z_1=z_2=\cdots=z_n=0$.

Solution: Lets call the sum of $k$-th powers $S_k$. Since all $S_k$-s are homogeneous polynomials we know that for any $\lambda\in\mathbb{C}$ the numbers $\lambda z_1,\lambda z_2,\ldots ,\lambda z_n$ will also fulfill these equations and call the corresponding polynomials $S_k(\lambda)$. Choose $\lambda'$, such that $|\lambda' z_i|<1$ for all $i=1,2,\ldots,n$. Now for $\lambda: \ |\lambda|\le |\lambda'|$ examine $$f(\lambda;z_i)=\sum_{k=1}^{\infty}\frac{(\lambda z_i)^k}{k}$$ This is a series for logarithm, so $$S:=\sum_{i=1}^{n}f(\lambda;z_i)=-\sum_{i=1}^{n}\log(1-\lambda z_i)=-\log((1- \lambda z_1)\cdots(1- \lambda z_n))$$
On the other hand $$S=\sum_{i=1}^{n}\sum_{k=1}^{\infty}\frac{(\lambda z_i)^k}{k}=\sum_{k=1}^{\infty}\frac{S_k(\lambda)}{k}=0$$
Thus we have $\log((1-\lambda z_1)\cdots(1-\lambda z_n))=0$ hence $$g(\lambda):=(1-\lambda z_1)\cdots(1-\lambda z_n)=1$$ for all $\lambda$ within a small disc around zero. This means that all coefficients of $g$ apart from the constant term are $0$. And these coefficients are exactly the elementary symmetric polynomials (in reverse order) of $z_1,z_2,\ldots,z_n$. So all of them are zeros - $z_1=z_2=\cdots=z_n=0$.

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