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(z^(n+1))-1 = (z-exp[2*Pi*i*(j+1)/(n+1)]) where 0<=j<=n

I tried doing it by induction which is easy for the 1st case with n = 1,I assume the case n-1 but then i get stuck with the last case.

How will i know that all the z's with a power smaller than n+1 will be zero since i know multiplying all the exp[2*Pi*i*(j+1)/(n+1)] from j=o to j= n+1 should give something like exp[2*Pi*i*(n(n+1))/(2(n+1)) +1] which will be one exp[2*Pi*i*(n(n+1))/(2(n+1)) +1 = cos(2*Pi*(n+2))+isin(2*Pi*(n+2)) and the all the z's will give me z^(n+1)

Can you please tell me what i can do or other ways of doing this so i can solve this