Trying to evaluate $\prod_{k=1}^{n-1}(1-e^{2k\pi i/n})$ for my complex analysis homework

For my complex analysis homework, I am trying to show that the integral of the real function $1/(1+x^n)$, for integer $n\ge2$, along the positive real line is $$\int_0^{\infty}\frac{dx}{1+x^n} = \frac{\pi}{n\sin(\pi/n)}.$$ The suggested approach is to consider the contour integral $$\int_{\gamma}f(z)dz = \int_{\gamma}\frac{dz}{1+z^n},$$ where $\gamma$ is a contour construced such that it traverses all of the positive real line, but only a single simple pole is included. So I started out by factoring the polynomial in the denominator, which has (distinct) roots $\omega_k=\exp(i[\pi+2k\pi]/n)$, so that $1+z^n=(z-\omega_0)(z-\omega_1)\cdots(z-\omega_{n-1})$. Using this, I attempt to find the residue at $z=\omega_0$ (which is the single pole included in $\gamma$ by construction) by $$\text{Res}_{\omega_0}f=\lim_{z\rightarrow\omega_0}\frac{1}{(z-\omega_0)\cdots(z-\omega_{n-1})}(z-\omega_0)= \frac{1}{(\omega_0-\omega_1)\cdots(\omega_0-\omega_{n-1})}.$$

I have found that for every $(\omega_0-\omega_k)=\exp(i\pi/n)-\exp(i\pi/n+2k\pi i/n)$, I can extract an $\omega_0$ from the $\omega_k$, yielding $(\omega_0-\omega_k)=\exp(i\pi/n)[1-\exp(2k\pi i/n)]$ such that we are left with $$\text{Res}_{\omega_0}f=\frac{1}{\prod_{k=1}^{n-1} e^{i\pi/n}[1-e^{2k\pi i/n}]}=\frac{1}{e^{i\pi(n-1)/n}\prod_{k=1}^{n-1} \left(1 - e^{2k\pi i /n}\right)}.$$

Now, since I know the final answer, I strongly believe that $$\prod_{k=1}^{n-1}(1-e^{2k\pi i/n} )=n,$$ but I am having a very hard time trying to evaluate this product. Any pointers would be greatly appreciated.