Does anyone know how to prove this formula?

$\displaystyle 2^{n-1}\prod_{k=0}^{n-1}\sin(x+\frac{k\pi}{n}) = \sin(nx)$

All suggestions, ideas are welcome.

Below you find my unsuccessfull attempt using complex numbers.

When you convert it to complex numbers the equality can be rewritten as

$\displaystyle 2^{n-1}\prod_{k=0}^{n-1}\frac{e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})}}{2i} = \frac{e^{inx}-e^{-inx}}{2i}$

$\displaystyle i^{-n+1}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}$

$\displaystyle e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}-e^{-i(x+\frac{k\pi}{n})} = e^{inx}-e^{-inx}$

The LHS of this equation contains terms in $\displaystyle e^{inx}$, $\displaystyle e^{i(n-1)x}$, $\displaystyle e^{i(n-2)x}$, ..., $\displaystyle e^{-inx}$.

Calculate the coefficient for each term.

$\displaystyle \underline{e^{inx}}$

$\displaystyle e^{i\frac{(-n+1)\pi}{2}}\frac{(-n+1)\pi}{2}\prod_{k=0}^{n-1}e^{i(x+\frac{k\pi}{n})}=e^{i\frac{(-n+1)\pi}{2}}e^{inx+i\sum_{k=0}^{n-1}\frac{k\pi}{n}}=e^{inx}$

$\displaystyle \underline{e^{-inx}}$

$\displaystyle e^{i\frac{(-n+1)\pi}{2}}\prod_{k=0}^{n-1}-e^{-i(x+\frac{k\pi}{n})}=(-1)^ne^{i(-n+1)\pi}e^{-inx}=(-1)^{n}e^{-inx} = -e^{-inx}$

Now we still have to prove that for -n<k<n the coefficient of $\displaystyle {e^{ikx}}$ equals 0 to conclude the proof. But I don't know how to do this.

Other proofs are also welcome.

Thanks!