# Prove that $\prod^{n}_{k=1}\sin\left(\frac{k\pi}{2n+1}\right) = \frac{\sqrt{2n+1}}{2^{n}}$ [duplicate]

Prove that $$\prod^{n}_{k=1}\sin\left(\frac{k\pi}{2n+1}\right) = \frac{\sqrt{2n+1}}{2^{n}}$$

$\bf{My\; Try::}$ Let $\displaystyle \cos \left(\frac{k\pi}{2n+1}\right)+i\sin \left(\frac{k\pi}{2n+1}\right)=e^{\frac{ik\pi}{2n+1}}=e^{ik \theta}\;,$ Where $\displaystyle \theta = \frac{\pi}{2n+1}$

So $$\prod^{n}_{k=1}\sin\left(\frac{k\pi}{2n+1}\right) = \Im \bigg(\prod^{n}_{k=1}e^{ik \theta}\bigg) = e^{i\theta}+e^{2i\theta}+.....+e^{ni\theta} = \frac{e^{i(n+1)\theta}-e^{i \theta}}{e^{i\theta}-1 }$$

Now How can I solve after that, Help me

Thanks