Using $\text{n}^{\text{th}}$ root of unity

$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$

Prove that 

$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$

$$ \begin{align*}
P & = \prod_{k=1}^{n-1}\sin(k\pi/n) \\
& = (2i)^{1-n}\prod_{k=1}^{n-1}(e^{ik\pi/n}-e^{-ik\pi/n}) \\
& = (2i)^{1-n} e^{-i \frac{n(n-1)}{2}\frac{\pi}{n}} \prod_{k=1}^{n-1}(e^{2ik\pi/n}-1) \\
& = (-2)^{1-n}\prod_{k=1}^{n-1}(\xi^k-1) \\
& = 2^{1-n}\prod_{k=1}^{n-1}(1-\xi^k) \\
\end{align*}
$$
where $\xi=e^{2i\pi/n}$.
Now note that $x^n-1=(x-1)\sum_{k=0}^{n-1}x^k$ and $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$.
Cancelling $(x-1)$ we have $\prod_{k=1}^{n-1} (x-\xi^k) =\sum_{k=0}^{n-1}x^k$.
Substituting $x=1$ we have $\prod_{k=1}^{n-1} (1-\xi^k)=n$. $$ \therefore \boxed{P=n2^{1-n}}$$

Edit:
In order to note that $x^n-1=\prod_{k=0}^{n-1} (x-\xi^k)$, note that $1,\xi,\dots,\xi^{n-1}$ are roots of $x^n-1$. Therefore by polynomial reminder theorem we have $x^n-1=Q(x) \prod_{k=0}^{n-1} (x-\xi^k)$. Comparing degrees we find $Q(x)$ has degree $0$. Comparing highest coefficients we conclude $Q(x)=1$.
Edit:
We may instead use the identity $\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2\sin(k\pi/n), k = 1, ..., n - 1,$ to establish immediately that $P \equiv \prod_{k=1}^{n-1}\sin(k\pi/n)= 2^{1-n}\prod_{k=1}^{n-1}\left\lvert 1 - e^{2ik\pi/n} \right\rvert = 2^{1 - n}\left\lvert \prod_{k=1}^{n-1}(1 - e^{2ik\pi/n}) \right\rvert$, and continue by applying the foregoing logic to the product to obtain $P=n2^{1-n}$.
