I am trying to prove an identity by Induction but I'm stuck and need some help. Here is the Statement $$\forall n\geq2,p(n)= \sum_{k=0}^{n-1}e^{i(\frac{2k\pi}{n})}=0.$$ I start with the base case $n=2.$ $$p(2)= \sum_{k=0}^{1}e^{i\big(\frac{2k\pi}{2}\big)}=\sum_{k=0}^{1}e^{i\big({k\pi}\big)}= \sum_{k=0}^{1}\big(\cos(k\pi)+i\sin(k\pi) \big)=1-1=0$$ Then we assume that $p(k\leq n)$ holds true and observe $p(k=n+1).$ $$ \sum_{k=0}^{n+1-1}e^{i(\frac{2k\pi}{n+1})}=\sum_{k=0}^{n} \cos \big(\frac{2k\pi}{n+1}\big)+i\sin\big(\frac{2k\pi}{n+1}\big) = \sum_{k=0}^{n-1} \bigg(\cos \big(\frac{2k\pi}{n+1}\big)+i\sin\big(\frac{2k\pi}{n+1}\big) \bigg)+\cos(\frac{2n\pi}{n+1})+i\sin(\frac{2n\pi}{n+1}).$$ I don't really know where I can go from here honestly, Is there some sort of identiy that I could use to make this easier?
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2Better use the geometric series. – Dietrich Burde Apr 17 '22 at 16:36
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2I don't think induction actually does much for you here, it's probably best to just try to prove it directly using geometric series – Stephen Donovan Apr 17 '22 at 16:37