I wonder if anyone has an idea about how to write
$$ \prod_{\substack{j=0\\ j\neq k}}^{n-1} ( e^{\frac{2\pi ik}{n}} - e^{\frac{2\pi ij}{n}})=n,\qquad k=0,1,...,n-1,\; j=0,1,....n-1$$
in a "general" polar form, i.e.
$$e^{\frac{2\pi ik}{n}} - e^{\frac{2\pi ij}{n}}$$
as $r (\cos\theta+i \sin\theta)$ so that I can make the proof by induction. Is it possible to do?
I'm very grateful for some help about this.
In general we have $e^{a + bi} = e^a e^{bi} = e^a(\cos(b) + i\sin(b))$ for all $a,b \in \mathbb{R}$. Thus for any complex number of the form $z = e^{i\theta}$ (with $\theta$ in radians a real number), we may write $z = \cos(\theta) + i\sin(\theta)$. In your case we have $e^{2\pi i k/n} = \cos(\frac{2 \pi k}{n}) + i\sin(\frac{2\pi k}{n})$. Note that $r = 1$ here as $a = 0$. Similarly with $e^{2\pi i j/n}$. I have a feeling that your proposed induction could get messy, but I have not done it out myself so this is pure speculation.
Hope this helped!
Does this help..? $$e^{i\frac{2 \pi k}{n}} - e^{i\frac{2 \pi j}{n}}=\cos \frac{2 \pi k}{n} + i\sin \frac{2 \pi k}{n} -\cos \frac{2 \pi j}{n}+i\sin \frac{2 \pi j}{n}= -2[\sin\frac{\pi (j+k)}{n}\sin\frac{\pi (j-k)}{n}] + 2i [\sin\frac{\pi (j+k)}{n} \cos\frac{\pi (j-k)}{n}]=2i \sin\frac{\pi (j+k)}{n} [\cos\frac{\pi (j-k)}{n} + i\sin\frac{\pi (j-k)}{n} ]$$
