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! 
 A: Hint 1: the $n$th roots of unity are the solutions to $x^n=1$. Letting $x:=re^{i\theta}$, we have $x^n=r^n e^{in\theta}$, which implies $r=1$ and $n\theta=2k\pi$ for any integer $k$.
Removing redundancies, the roots of unity are...?

  $e^{2\pi i/n},e^{2\cdot 2\pi i/n},\ldots,e^{(n-1) \cdot 2\pi i/n},1$.

Hint 2:
Taking the product of the roots of unity gives...?

$$e^{2\pi i \cdot (1+2+3+\cdots (n-1))/n}=e^{2\pi i(n-1)/2} = \begin{cases}1 & \text{$n$ odd}\\ -1 & \text{$n$ even}\end{cases} = (-1)^{n+1}$$

A: 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 
$$\prod_{k=0}^n r_n =  e^{i(n-1)\pi} =   \left( e^{i\pi} \right)^{(n-1)} = (-1)^{(n-1)}  $$
A: Hint: an $n$th root of unity has the form 
$$
\exp\left({\frac{2\pi ik}{n}}\right)
$$
for $1\leq k\leq n$.
A: 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.
