Product of roots of unity using e^xi Find the product of the $n\ n^{th}$ roots of 1 in terms of n. The answer is $(-1)^{n+1}$ but why? Prove using e^xi notation please! 

The $n $ roots of unity are 
$$ r_k = e^{i \frac{2 \pi k }{n}} \, \text{ for }k=0 ...n-1 $$
so 
$$\prod_{k=0}^n r_n =   \prod_{k=0}^n e^{i \frac{2 \pi k }{n}} =e^{i \frac{2 \pi X}{n}}  $$
where $$X = \sum_{k=0}^{n-1} k = \frac{n(n-1)}{2} $$ 
so $\frac{2 \pi X}{n} = (n-1)\pi$
leaving 
$$\prod_{k=0}^n r_n =  e^{i(n-1)\pi} =   \left( e^{i\pi} \right)^{(n-1)} = (-1)^{(n-1)}  $$
