The partial fraction expansion of $\frac{1}{x^n - 1}$.

If $n$ is an integer, is there a nice way to write the partial fraction expansion of $\frac{1}{x^n - 1}$? I figure that if $\zeta$ is the $n$-th root of unity, then for some coefficients $a_0, a_1, \ldots, a_{n-1}$ we may write

$$ \frac1{x^n - 1} = \frac{a_0}{x - 1} + \frac{a_1}{x - \zeta} + \frac{a_2}{x - \zeta^2} + \ldots + \frac{a_{n-1}}{x - \zeta^{n-1}}. $$

Then for $0 \leq i \leq n -1$, $$ a_i = \lim_{x \to \zeta^i} \frac{x - \zeta^i}{x^n - 1} = \frac1{(\zeta^{i} - 1) \cdots (\zeta^i - \zeta^{i-1}) (\zeta^i - \zeta^{i + 1}) \cdots (\zeta^i - \zeta^{n-1})}. $$ Is there a simpler expression for this and if so, how could I see it easily?

Use l'Hopital's rule. $a_i = \lim_{x \to \zeta^i} \frac{1}{nx^{n-1}} = \frac{\zeta^i}{n}$. Note that the identity I am claiming is true is equivalent to the identity

$$\sum_{i=0}^{n-1} (\zeta^i)^k = \begin{cases} n \text{ if } n | k \\\ 0 \text{ otherwise} \end{cases}$$

by taking the generating function of both sides. More generally, the coefficient of $\frac{1}{x - a}$ in the partial fraction decomposition of $\frac{P(x)}{Q(x)}$ is (as long as $a$ is a simple root of $Q$)

$$\lim_{x \to a} \frac{(x - a) P(x)}{Q(x)} = \frac{P(a)}{Q'(a)}.$$

This formula is tremendously useful (for example I recently used it on this math.SE question) but does not seem to be widely known.