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! 

Let $x_0, x_1, \ldots, x_{n-1}$ denote the $n$ $n$-th roots of unity. All of these are roots of the polynomial $x^n-1$, that is.
$$x^n-1 = (x-x_0)(x-x_1)(x-x_2)\cdots (x-x_{n-1}).$$
The constant term on the right side is the product
$(-x_0)(-x_1)(-x_2)\cdots (-x_{n-1}) = (-1)^nx_0x_1\cdots x_{n-1}$
while the constant term on the left side is $-1$.
Hence,
\begin{align}
(-1)^nx_0x_1\cdots x_{n-1} &= -1\\
(-1)^n \cdot (-1)^n x_0x_1\cdots x_{n-1} &= (-1)^n\cdot(-1)\\
(-1)^{2n}x_0x_1\cdots x_{n-1} &= (-1)^{n+1}\\
x_0x_1\cdots x_{n-1} &= (-1)^{n+1}
\end{align}
Feel free to substitute $e^{(i2\pi/n)k}$ for $x_k$, $0 \leq k < n$
if you think it improves the proof. But the more general form
discussed here makes it applicable to $n$-th roots of unity in
other fields, e.g. finite fields.
