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Yes, assuming we are in $\mathbb{C}$. A first consequence is that all the symmetric functions of $(c_1,\dots,c_n)$ are zero (see e.g. this wiki article). But this can be written as an identity of polynomials $$x^n=\prod_{j=0}^n(x-c_j)$$ which obviously implies $c_j=0$ for all $j$.