Simplifying $\prod\limits_{k\neq j=0}^{n-1}\frac1{\lambda_{n,k}-\lambda_{n,j}}$ for $\lambda_{n,k}=\exp\frac{i\pi(2k+1)}{n}$

Defining the polynomial \begin{align} P(x)&=x^n+1\\ &=\prod_{j=0}^{n-1}\left( x- \lambda_{n,j}\right) \end{align} we can express its derivative at $x=\lambda_{n,k}$ as: \begin{align} P'(\lambda_{n,k})&=\prod_{k\neq j=0}^{n-1}\left( \lambda_{n,k}-\lambda_{n,j} \right)\\ &=\frac{1}{\Gamma_{n,k}} \end{align} But we have also $P'(x)=nx^{n-1}=n\tfrac{x^n}{x}$. Thus, as $\left(\lambda_{n,k} \right)^n=-1$, \begin{equation} P'(\lambda_{n,k})=n\frac{-1}{\lambda_{n,k}} \end{equation} Finally, \begin{equation} \Gamma_{n,k}=-\frac{\lambda_{n,k}}{n} \end{equation} This trick comes rather naturally if the integral is evaluated by the residue method, for the function $f(z)=(1+z^n)^{-1}\ln\left(\tfrac z{1-z}\right)$ along the keyhole contour.