Simplifying $\prod\limits_{k\neq j=0}^{n-1}\frac1{\lambda_{n,k}-\lambda_{n,j}}$ for $\lambda_{n,k}=\exp\frac{i\pi(2k+1)}{n}$ I have been able to show that for $n\in\Bbb N_{\geq2}$ $$\phi(n)=\int_0^1\frac{dx}{x^n+1}=\sum_{k=0}^{n-1}\Gamma_{n,k}\log\frac{\lambda_{n,k}-1}{\lambda_{n,k}}$$
Where $$\lambda_{n,k}=\exp\frac{i\pi(2k+1)}{n}$$
And $$\Gamma_{n,k}=\prod_{k\neq j=0}^{n-1}\frac1{\lambda_{n,k}-\lambda_{n,j}}$$
And I was wondering: how do we simplify $\Gamma_{n,k}$ to ease the manual calculation of $\phi(n)$ values. The integral is always real, so I am sure there is a major way we can simplify $\Gamma_{n,k}$, but I have been so far unable to find it. I do suspect however that the product $$P_n=\prod_{k=0}^{n-1}\Gamma_{n,k}$$
May play a significant role in finding the simplification I seek. 

For those interested, a proof. 
Note that $x^n+1$ bay be factored as
$$x^n+1=\prod_{k=0}^{n-1}(x-\lambda_{n,k})$$
Hence $$\phi(n)=\int_0^1\prod_{k=0}^{n-1}\frac1{x-\lambda_{n,k}}dx$$
Then define $\Gamma_{n,k}$ by saying that
$$\prod_{k=0}^{n-1}\frac1{x-\lambda_{n,k}}=\sum_{k=0}^{n-1}\frac{\Gamma_{n,k}}{x-\lambda_{n,k}}$$
Multiplying both sides by $\prod_{j=0}^{n-1}(x-\lambda_{n,j})$:
$$1=\sum_{k=0}^{n-1}\frac{\Gamma_{n,k}}{x-\lambda_{n,k}}\prod_{j=0}^{n-1}(x-\lambda_{n,j})$$
$$1=\sum_{k=0}^{n-1}\Gamma_{n,k}\prod_{k\neq j=0}^{n-1}(x-\lambda_{n,j})$$
So for any integer $0\leq m\leq n-1$ we may plug in $x=\lambda_{n,m}$ and simplify to get 
$$\Gamma_{n,m}=\prod_{m\neq j=0}^{n-1}\frac1{\lambda_{n,m}-\lambda_{n,j}}$$
And our result follows directly. 
Perhaps another motivation for easing manual calculation of this product would be that $$\sum_{k=0}^{\infty}\frac{(-1)^k}{nk+1}=\phi(n)$$
Which brings about a plethora of interesting closed forms.

Edit: A little progress
We define $$c_{n,j}=\operatorname{Re}\lambda_{n,j}=\cos\frac{\pi(2j+1)}{n}$$
And $$s_{n,j}=\operatorname{Im}\lambda_{n,j}=\sin\frac{\pi(2j+1)}{n}$$
So $$\log\frac{\lambda_{n,k}-1}{\lambda_{n,k}}=\log\left(1-\lambda_{n,k}^{-1}\right)=\log\left(1-c_{n,k}+is_{n,k}\right)$$
And we also see that 
$$\begin{align}
\prod_{k\neq j=0}^{n-1}\frac1{\lambda_{n,k}-\lambda_{n,j}}&=\prod_{k\neq j=0}^{n-1}\frac1{e^{i\pi(2k+1)/n}-e^{i\pi(2j+1)/n}}\\
&=\prod_{k\neq j=0}^{n-1}\frac{e^{-i\pi(2k+1)/n}}{1-e^{i\pi(2j-2k)/n}}\\
&=e^{i(2k+1)(2-n)/n}\prod_{k\neq j=0}^{n-1}\frac12\left(1+i\cot\frac{\pi(j-k)}n\right)\\
\Gamma_{n,k}&=\frac{\lambda_{n,k}^{2-n}}{2^{n-2}}\prod_{k\neq j=0}^{n-1}\left(1+i\cot\frac{\pi(j-k)}n\right)
\end{align}$$
But the remaining product I do not know how to deal with.
 A: 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.
A: In fact, if we may write a function $f$ as a product over its roots, i.e. 
$$f(x)\equiv\prod_{f(\omega)=0}(x-\omega)$$
where each root $\omega$ contributes exactly one term, then we may also write 
$$\frac1{f(x)}=\sum_{f(\omega)=0}\frac{b(\omega)}{x-\omega}$$
for some coefficients $b(\omega)$ which we can show to be $$b(\omega)=\prod_{{f(\alpha)=0}\atop{\alpha\ne \omega}}\frac1{\omega-\alpha}.$$
At the same time however, we have that the equality
$$f'(x)=\sum_{f(\omega)=0}\frac{f(x)}{x-\omega}=\sum_{f(\omega)=0}\prod_{{f(\alpha)=0}\atop{\alpha\ne \omega}}(x-\alpha)$$
holds under the assumption that $f(x)=0\Rightarrow f'(x)\ne0$.  
So for any root $\phi$ we plug in $x=\phi$ to see that
$$f'(\phi)=\prod_{{f(\alpha)=0}\atop{\alpha\ne \phi}}(\phi-\alpha)$$
which implies that $$b(\omega)=\frac{1}{f'(\omega)}$$
and 
$$\frac1{f(x)}=\sum_{f(\omega)=0}\frac1{(x-\omega)f'(\omega)}.$$
Then @PaulEnta's results are easily derived form there.
