$\lim_{n \to +\infty}2^n\sin(2^{-n})$ Software tells me that $\lim_{n \to +\infty}2^n\sin(2^{-n}) = 1$. 
Can someone give me a hint on where to start to actually show this is true?
I was working on the following problem:
$$ \prod_{n=0}^\infty\cos(2^{-n})$$
And I used,
$$\sin(1) = 2\sin(2^{-1})\cos(2^{-1}) $$
$$\sin(1) = 2^{n}\sin(2^{-n})\cos(2^{-n}) \cdots \cos(2^{-1})$$
$$\sin(1)\cos(1) = 2^{n}\sin(2^{-n})\cos(2^{-n}) \cdots \cos(2^{-1})\cos(1)$$
So if $$\lim_{n \to +\infty}2^nsin(2^{-n}) = 1$$
then,
$$\sin(1)\cos(1) = \prod_{n=0}^\infty\cos(2^{-n})$$.
I feel like this should be relatively simple, but I'm pulling a blank.

Hint $\lim_{x\to\infty}x\sin {1\over x}=\lim_{y\to 0}{\sin y\over y}=1$ by taking $y={1\over x}$
