What does $\sin x \cdot \sin 2x \cdot \sin 3x \cdot ... \cdot \sin nx$ equal to? 
Problem: Find a general formula for:
$$A = \sin x \cdot \sin 2x \cdot \sin 3x \cdot ... \cdot \sin nx$$

How do I come across with this? Or is there a way to simplify this? I tried to at least guess the formula but it seems complicated.

Edit: From the comments, and some thoughts from me, I suspect that the answer to the above could be

$$A = 2^{-n} \cdot i^n \cdot \prod_{k=1}^{n} \, \left(e^{-kix} - e^{kix}\right)$$

*

*Is this formula correct? (Can it be further simplified?)


*If yes, how do I prove it?
 A: Yes, correct.
Euler's famous identity:  $$e^{i\phi} = \cos \phi + i \sin \phi $$ can also be written (replace $\phi$ with $-\phi$)
$$e^{-i\phi} = \cos \phi - i \sin \phi $$
Now take their difference these to see that $$\sin \phi = \frac{ e^{i \phi } - e^{-i\phi} }{2i}$$ which leads to your expression $(\phi = kx, \quad k=1,2,\cdots,n).$
Similarly, adding leads to $$\cos \phi = \frac{ e^{i \phi } + e^{-i\phi} }{2}.$$
You could multiply these binomials and see where that leads.  I don't know if you get something 'simpler' other than putting the result back into a form involving only sines and cosines.
UPDATE
Here are the results when applying the expression for $n=2$ and $n=3$ and then factoring to get the terms to involve only $\cos x$ and $\sin x$:
$$\sin x \cdot \sin 2x=\frac{1}{2} \cos x - \frac{1}{2} \cos^3 x + \frac{3}{2} \cos x \sin^2 x.$$
$$\sin x \cdot \sin 2x \cdot \sin 3x
=\frac{1}{2} \cos x \sin x
+\cos^3 x  \sin x
- \frac{3}{2} \cos^5~x \sin~x
-~\cos~x~\sin^3~x
+~5~\cos^3~x~\sin^3~x
-~\frac{3}{2}~\cos~x~\sin^5~x.$$
To get these forms, Let $p=e^{ix}$ and $q=e^{-ix}$ to
obtain a polynomial in $p$ and $q$.  Now factor to get terms involving only $p-q$ and $p+q$.  These terms are (up to a constant) $\sin x$ and $\cos x$.
Simpler than your expression for $A$? 
