Note that $\sin(\theta)=(2i)^{-1}(e^{i\theta}-e^{-i\theta})=(2i)^{-1}e^{-i\theta}(e^{i\cdot 2\theta}-1)$ for all $\theta\in\mathbb R$. Set $\theta=\pi/n$. Then$$\prod_{k=1}^{n-1}\sin(k\pi/n)=\prod_{k=1}^{n-1}\sin(k\theta)=(2i)^{-(n-1)}\prod_{k=1}^{n-1}e^{-i\cdot k\theta}\cdot\prod_{k=1}^{n-1}(e^{i\cdot 2k\theta}-1).$$Now,$$\prod_{k=1}^{n-1}e^{-i\cdot k\theta}=e^{-i\cdot\sum_{k=1}^{n-1}k\pi/n}=e^{-i\cdot (n-1)\pi/2}=(-i)^{n-1}.$$Set $\zeta=e^{i\cdot 2\theta}=e^{i\cdot 2\pi/n}$. Then $\zeta, \zeta^2, \dots, \zeta^{n-1}$, together with 1 are exactly the nth roots of unity (i.e. the roots of the polynomial $x^n-1$). Therefore, we have the factorization$$x^n-1=(x-1)\cdot\prod_{k=1}^{n-1}(x-\zeta^k),$$and so$$x^{n-1}+\dots+x+1=\prod_{k=1}^{n-1}(x-\zeta^k).$$With $x=1$, we have$$n=\prod_{k=1}^{n-1}(1-\zeta^k)=(-1)^{n-1}\prod_{k=1}^{n-1}(e^{i\cdot 2k\theta}-1).$$Putting them together, we get$$\prod_{k=1}^{n-1}\sin(k\pi/n)=(2i)^{-(n-1)}\cdot(-i)^{n-1}\cdot n/(-1)^{n-1}=n/2^{n-1}$$as desired.