Let $ f(z) = \frac{z}{z^n - 1} $. You're looking for $ \oint_{\gamma} f(z) dz $ taken over a contour that is outside of the unit circle, loosely speaking. We will use the Residue Theorem to compute the integral. Define $ \omega_k = e^{2 \pi i k / n} $ for $ k = 0, \dots, n-1 $ to be the roots of unity, which are also the poles of $ f(z) $. We calculate the residues using the factorization $ z^n - 1 = \prod_{k=0}^{n-1} (z-\omega_k) $ as follows:
$ R_k = \lim_{z\to \omega_k} (z-\omega_k)f(z) = \omega_k \prod_{j \ne k} (\omega_k - \omega_j)^{-1} = \omega_k \prod_{l = 1}^{n-1} \left[ \omega_k^{-1} ( 1 - \omega_l)^{-1} \right] = \omega_k^2 R_0 .$
Note that the above uses $ \omega_k^{-n} = 1 $ and the index substitution $ l = j-k $. We could use the geometric formula to calculate $ R_0 = 1/(n-1) $, but it's actually unnecessary because we know that $ \sum \omega_k^2 = 0 $ (for n > 2 ), since it is the $ z^2 $ coefficient of the polynomial $ z^n-1 $. Hence the integral vanishes.