Rigorous proof of an infinite product. I'll give a proof of the following expansion:
$$\frac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \frac{x}{2^i}$$
$${\sin x} = 2 \cos \frac{x}{2}\sin  \frac{x}{2}$$
$${\sin x} = 2^2 \cos \frac{x}{2}\cos  \frac{x}{4}\sin  \frac{x}{4}$$
$$ {\sin x} = 2^3 \cos \frac{x}{2} \cos \frac{x}{4} \cos \frac{x}{8} \sin\frac{x}{8} $$
$$ {\sin x} = 2^k \cos \frac{x}{2} \cos \frac{x}{4} \cdots\cos \frac{x}{2^k} \sin\frac{x}{2^k} $$
$${\sin x} = 2^k \sin\frac{x}{2^k} \prod_{i=1}^{k} \cos \frac{x}{2^i}  $$
$$ \frac{\sin x}{x} = \frac{\sin \displaystyle \frac{x}{2^k}}{\displaystyle \frac{x}{2^k}} \prod_{i=1}^{k} \cos \frac{x}{2^i}  $$
Let $k \to \infty$, then $\displaystyle \frac{x}{2^k} \to 0$
$$ \frac{\sin x}{x} = \prod_{i=1}^{\infty} \cos \frac{x}{2^i}  $$
Is there any extra observation needed to make the proof complete?
I guess since $$\cos \frac{x}{2^i} \to 1$$
the convergence is not at stake.

Your expression 
$$ \frac{\sin x}{x} = \frac{\sin  \frac{x}{2^k}}{ \frac{x}{2^k}} \prod_{i=1}^{k} \cos \frac{x}{2^i}  $$
is correct. Maybe we should separate out the very special case $x=0$, and from then on assume that $x\ne 0$. For $x=0$,  $\frac{\sin x}{x}$ is formally undefined, but it is natural to set it equal to $1$. Then the formula works. If we are feeling in a very formal mood, we should  prove the  correctness of your expression by induction on $k$. However, I think that would be overkill.
We want to find the limit of the expression on the right as $k\to \infty$. As you observed, $\frac{x}{2^k}\to 0$  as $k\to \infty$. That certainly does not need proof. However, it is necessary to observe that since 
$$\lim_{t=0}\frac{\sin t}{t}=1,$$
we have 
$$\lim_{k\to\infty} \frac{\sin  \frac{x}{2^k}}{ \frac{x}{2^k}}=1.$$
There will unfortunately be some special cases that require special treatment. We deal first with the "general" case when $x$ is not an integer multiple of $\pi$.
Then your expression can be rewritten as 
$$\frac{\sin x}{x}\frac{\frac{x}{2^k}}{\sin\frac{x}{2^k}}=\prod_{i=1}^{k} \cos \frac{x}{2^i}.$$
Since 
$$\lim_{k\to\infty}\frac{\sin x}{x}\frac{\frac{x}{2^k}}{\sin\frac{x}{2^k}}$$
exists and is equal to $\frac{\sin x}{x}$, we conclude that 
$$\lim_{k\to\infty} \prod_{i=1}^{k} \cos \frac{x}{2^i}$$
also exists and is equal to $\frac{\sin x}{x}$.
If $x\ne 0$ is an integer multiple of $\pi$, we have to choose $k$ large enough so that $\sin\frac{x}{2^k}\ne 0$, else when we rewrite your expression, we might be dividing by $0$. Minor point. For such $x\ne 0$, $\frac{\sin x}{x}=0$, and one of the cosines is $0$. So, in that case also, apart from a technicality discussed below, the formula looks correct. 
Technical remark:  In the formal definition of an infinite product, we say that 
$$\prod_{i=1}^\infty a_i$$
converges if 
$$\lim_{k\to\infty}\prod_{i=1}^k a_i$$
exists and is not equal to $0$.  So technically when $x$ is an integer multiple of $\pi$, the infinite product does not converge!
