$\lim_{n \to +\infty}2^n\sin(2^{-n})$ 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.

Perform a substitution $2^n=t$ then rewrite the limit.
The term becomes $t\sin(1/t)$ with t→ infinity.
If you then perform another sub, say $t=1/v$, where v goes to zero, you will recognize a familiar standard limit which is 1
