Proving: $\cos A \cdot \cos 2A \cdot \cos 2^{2}A \cdot \cos 2^{3}A ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin A } $ $$\cos A \cdot \cos 2A  \cdot \cos 2^{2}A  \cdot \cos 2^{3}A  ... \cos 2^{n-1}A = \frac { \sin 2^n A}{ 2^n \sin  A } $$
I am very much inquisitive to see how this trigonometrical identity can be proved.
PS:I am not much of interested about an inductive proof.

You can prove it by induction since
$\sin t = 2 \cos \dfrac{t}{2}\sin \dfrac{t}{2}$
$\sin t = 2^2 \cos \dfrac{t}{2}\cos \dfrac{t}{4}\sin \dfrac{t}{4}$
$\sin t = {2^3}\cos \dfrac{t}{2}\cos \dfrac{t}{4}\cos \dfrac{t}{8}\sin \dfrac{t}{8}$
So we conjecture:
$$\sin t = 2^n\sin\dfrac{t}{2^n} \prod_{k=1}^{n} \cos\dfrac{t}{2^k} $$
It is true for $n=1$
$$\sin t = 2^1\sin\dfrac{t}{2^1} \prod_{k=1}^{1} \cos\dfrac{t}{2^1} = 2 \cos \dfrac{t}{2}\sin \dfrac{t}{2}  $$
But then for $n+1$ we get
$$\sin t = 2^{n+1}\sin\dfrac{t}{2^{n+1}} \prod_{k=1}^{n+1} \cos\dfrac{t}{2^{k}} $$
$$\sin t = {2^n}2\sin \frac{t}{{{2^{n + 1}}}}\cos \frac{t}{{{2^{n + 1}}}}\prod\limits_{k = 1}^n {\cos } \frac{t}{{{2^k}}}$$
$$\sin t = {2^n}\sin \frac{t}{{{2^n}}}\prod\limits_{k = 1}^n {\cos } \frac{t}{{{2^k}}}$$
