Product of the n roots of the unity can be $(-1)^{n-1}$

Question

I need to prove that the product of the n roots of the unity can be $(-1)^{n-1}$.

If i make $1^n=z$, z can be $ cis(\dfrac{k \cdot 2\Pi}{n})$ with k=0, ... , n-1

Now i need to prove that $\prod_0^{n-1}cis(\dfrac{k \cdot 2\Pi}{n}) = (-1)^{n-1}$

Answer 1

$$\prod_{k=0}^{n-1}cis(\dfrac{k \cdot 2\pi}{n})\\ = \prod_{k=0}^{n-1}e^(\dfrac{k \cdot 2\pi i}{n}) \\= e^{\sum_{k=0}^{n-1}\frac{2k\pi i}{n}} \\ = e^{\frac{2\pi i}{n}\sum_{k=0}^{n-1}k} \\ = e^{\frac{2\pi i}{n}(\frac{1}{2}(n)(n-1))} \\ = e^{\pi i{(n-1)}} \\ =(e^{\pi i})^{(n-1)} \\ = (-1)^{(n-1)}$$

Answer 2

Hint: Use the fact that the $n$'th roots of unity are given by $$ \zeta_n^k=e^{2\pi i \frac kn} $$ and multiply those.

Answer 3

Hint: Apply Vieta's formula to the polynomial $z^n-1.$