Let $x_1, \ldots,x_n$ be complex numbers such that for any $k$, $ \sum_{i=0}^n x_i^k = 0.$
I'd like to show that this implies $x_1 = x_2 = \cdots = x_n = 0.$
I was suggested to use this strategy. Suppose $|x_1| \geq |x_2| \geq \cdots \geq |x_n|.$ Then $ \lim_{k\to \infty} \left(\sum_{i=0}^n x_i^k \right)/x_1 = d$ where $d$ is the total of all numbers with maximal absolute value.
This would then imply that for large enough $k$ the sum from the proposition is not zero.
What I was wondering is, is there a more straightforward way of showing this?