Of course, $\sin x \prod_{k=0}^{23} \cos 2^k x = \dfrac {1} {2^{24}} \sin 2^{24}x$. For $x=56$, we have $2^{24} \cdot 56 \equiv 56 \pmod{360}$, since equivalent to $7(2^{24} - 1) \equiv 0 \pmod{45}$, and $\varphi(45) = 24$. Thus $\sin 2^{24}x = \sin x$ and we're done.

We use $\cos{2a}=2\sin{a}\cos{a}$, thus \begin{align*} \frac{\sin(2 \cdot 56^{\circ})}{2 \sin(56^{\circ})}\cdot \frac{\sin(4 \cdot 56^{\circ})}{2 \sin(2\cdot 56^{\circ})}\cdot \dots \cdot \frac{\sin(2^{24} \cdot 56^{\circ})}{2 \sin(2^{23}\cdot 56^{\circ})}=\frac{\sin(2^{24}\cdot 56^{\circ})}{2^{24}\cdot \sin(56^{\circ})} \end{align*}Note that $360 \mid (2^{24}-1)\cdot 56$ as $\phi(45)=24$, we have $\sin(56^{\circ})=\sin(2^{24}\cdot 56^{\circ})$. We are done.

We use "Vincent Huang Bashing". Let $z=e^{\frac{14\pi i}{45}}$. Using the key cosine result of this method, it reduces to showing $$\prod_{k=0}^{23} \frac{1}{2}\left(z^{2^k}+\frac{1}{z^{2^k}}\right) = \frac{1}{2^{24}}\prod_{k=0}^{23} \left(z^{2^k}+\frac{1}{z^{2^k}}\right) = \frac{1}{2^{24}} \rightarrow \prod_{k=0}^{23} \left(z^{2^k}+\frac{1}{z^{2^k}}\right)=1,$$which is clear by using $z^{45}=1$. $\blacksquare$