$$(c_i-c_j)^k = \sum_{h=0}^k \binom{k}{h} (-1)^{k-h}c_i^h c_j^{k-h}$$
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 $c_i^h$ are independent because they form a Vandermonde matrix.