Putting roots together

With the binomial theorem, the $k$ th term is equal to $\sum_{\ell = 0}^7 \binom 7 \ell \zeta_7^{(\ell-3)k};$ when summing over the values of $k$ , the exponents $(\ell - 3)k$ traverse all values $0,\cdots,6$ modulo 7; thus they contribute nothing, since $\sum_{j=0}^6 \zeta_7^j = 0.$ However, if $\ell = 3$ then $\sum_{k=0}^6 \zeta_7^{(\ell-3)k} = \sum_{k=0}^6 1 = 7$ . Thus the expression evaluates to $7\cdot \binom 7 3 = \boxed{245}.$ In general, $\sum_{k=0}^{n-1} \zeta_n^{-rk} (1 + \zeta_n^k)^n = n\binom n r.$