$$\frac{\sin x}{x} = \prod_{k=0}^\infty \sec(2^kx)$$
I'm unsure, so I'm looking for confirmation, this is my proof for the 2-adics:
Start with the common identity, $$\sin x (2 \cos x) = \sin (2x)$$ It telescopes to arbitrary values,
$$\sin x \left(2^n\prod_{k=0}^{n-1} \cos(2^k x) \right)= \sin(2^n x)$$ Divide both sides by $2^n$ and then take the limit as $n \to \infty$
$$\sin x \prod_{k=0}^\infty \cos(2^k x) = x$$
Divide to get the form:
$$\frac{\sin x}{x} = \prod_{k=0}^\infty \sec(2^kx)$$