I've stumbled upon this product of sines: $$ \sin{x}\sin{2x}\sin{3x}...\sin{nx}$$ I have to find a general formula for it. So far I tried to use the complex definition for the sine: $$\sin{x}=\frac{e^{ix}-e^{-ix}}{2i}$$
$$\sin{x}\sin{2x}\sin{3x}...\sin{nx}=\prod_{k=1}^{n}\frac{e^{kix}-e^{-kix}}{2i}=\frac{1}{(2i)^n}\prod_{k=1}^{n}e^{-kix}(e^{2kix}-1) \\ =\frac{1}{(2i)^n}\prod_{k=1}^{n}e^{-kix}\prod_{k=1}^{n}(e^{2kix}-1)$$ I got stuck when it came to this product:
$$ \prod_{k=1}^{n} {e^{2kix}-1}$$
What would you suggest? What should I do next?