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
][1]). 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 $j$.

[1]:http://en.wikipedia.org/wiki/Power_sum_symmetric_polynomial