$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$
and each summand is a rank-1 matrix (since it's a function of $i$ times a function of $j$). To prove that the rank is not lower than that, consider that the vectors $\mathbf{v}_h = (c_i^h)_{i=1}^n$ are independent because they form a Vandermonde matrix, and so are the $\mathbf{w}_h = (c_j^h)_{j=1}^n$.