Find $\prod_{j = 1, j \ne k}^{n} \left( e^{\frac{(2k - 1)i \pi}{n}} - e^{\frac{(2j - 1)i \pi}{n}} \right)$ I'm trying to use fractions decomposition to integrate
$$
\frac{1}{x^n + 1}
$$
As roots of the denominator are nth roots of $-1$, the denominator can be written as 
$$
\prod_{j = 1}^{n} \left( x - e^{\frac{(2j - 1)i \pi}{n}} \right)
$$
And the fractions decomposition can be written as
$$
\sum_{j = 1}^{n} \frac{A_j}{x - r_j}, \text{where $r_j$ is jth root of denominator ($e^{\frac{(2j - 1)i \pi}{n}}$)}
$$
Having said that, I need to find $A_j$. To do so, let's look at the numerator of a resulting fraction, it has the form
$$
\sum_{k = 1}^{n} A_k \prod_{j = 1, j \neq k}^{n} (x - r_j)
$$
At point $x = r_k$ it should be equal to 1, so we get 
$$
A_k \prod_{j = 1, j \neq k}^{n} (x - r_j) = 1
$$
During my research, I found a statement without proof, that the product is equal to $-n e^{-\frac{(2k - 1)i \pi}{n}}$ so
$$
\prod_{j = 1, j != k}^{n} \left( e^{\frac{(2k - 1)i \pi}{n}} - e^{\frac{(2j - 1)i \pi}{n}} \right) = -n e^{-\frac{(2k - 1)i \pi}{n}} \quad (1)
$$
But I failed to prove it yet. I've rewritten the product as 
$$
\left( e^{\frac{(2k - 1)i \pi}{n}} \right)^n \prod_{j = 1, j != k}^{n} \left( 1 - e^{\frac{2i \pi (k - j)}{n}} \right)
$$
And I see that if $e^{\frac{2i \pi (k - j)}{n}}$ are the roots of some polynomial $P(x)$, than I can easily calculate the product as $P(1)$, but I can't construct such $P(x)$. Is there any easier way to prove the statement (1)?
 A: Let $\zeta:=e^{\frac{\pi i}{n}}$ so that your product is
$$\prod_{\substack{j=1\\j\neq k}}^n\left(\zeta^{2k-1}-\zeta^{2j-1}\right).$$
As you already note, this product can be rewritten as
$$\prod_{\substack{j=1\\j\neq k}}^n(\zeta^{2k-1}-\zeta^{2j-1})
=(\zeta^{2k-1})^{n-1}\prod_{\substack{j=1\\j\neq k}}^n
\left(1-\zeta^{2(j-k)}\right).$$
A change of variables $i:=j-k$ and the fact that $\zeta^n=-1$ show that
$$(\zeta^{2k-1})^{n-1}\prod_{\substack{j=1\\j\neq k}}^n\left(1-\zeta^{2(j-k)}\right)
=-\zeta^{1-2k}\prod_{\substack{i=1-k\\i\neq0}}^{n-k}\left(1-\zeta^{2i}\right).$$
Note the following convenient identity of polynomials:
\begin{eqnarray*}
\prod_{\substack{i=1-k\\i\neq0}}^{n-k}
\left(X^2-\zeta^{2i}\right)
&=&\prod_{\substack{i=1-k\\i\neq0}}^{n-k}
(X-\zeta^i)(X+\zeta^i)
=\prod_{\substack{i=1-k\\i\neq0}}^{n-k}
(X-\zeta^i)(X-\zeta^{i+n})\\
&=&\prod_{\substack{i=1-k\\i\neq0\\i\neq n}}^{2n-k}
(X-\zeta^i)
=\prod_{\substack{i=1\\i\neq n}}^{2n}
(X-\zeta^i)
\end{eqnarray*}
where the latter equality holds because $\zeta^{2n}=1$. Of course this product is familiar;
$$\prod_{\substack{i=1\\i\neq n}}^{2n}(X-\zeta^i)
=\frac{X^{2n}-1}{X^2-1}
=\sum_{l=0}^{n-1}X^{2l}.$$
Plugging in $X=1$ shows that
$$\prod_{\substack{i=1-k\\i\neq0}}^{n-k}
\left(1-\zeta^{2i}\right)=n,$$
and hence your original product equals $-\zeta^{1-2k}n$, as you already found.
A: As the equation $$1+x^n=0$$ has $n$ distinct roots:
$$
\zeta_k=e^{i\frac{2k-1}{n}\pi},\quad k=1\dots n,
$$
the inverse of the polynomial can be written as:
$$
\frac{1}{1+x^n}=\sum_{k=1}^n\frac{c_k}{x-\zeta_k}.
$$
Multiplying both sides of the equation by $(x-\zeta_k)$ and taking the limit $x\to\zeta_k$ one obtains:
$$
\begin{array}{}
c_k&=\lim_{x\to\zeta_k}\frac{x-\zeta_k}{1+x^n}\\
&=\lim_{x\to\zeta_k}\frac{1}{nx^{n-1}}\\
&=\frac{1}{n\zeta_k^{n-1}}
=\frac{\zeta_k}{n\zeta_k^{n}}=-\frac{\zeta_k}{n}.\end{array}
$$
where L'Hospital rule was apllied to obtain the second equality.
PS. The product you asked about is the inverse of $c_k$.
