Questions follow a complex life cycle on Math StackExchange. Some questions are deleted, because they don't meet the community standards. Nonetheless, some of those deleted questions are interesting and non-trivial. In this post, I will cover one of those questions (you may not be able to see it, without sufficient privilege). I spent a few good hours to spot the pattern, so it won't feel like I wasted that time :) ... and you might like the question as well.
Is it true that $$\prod\limits_{k=1}^{2^{1999}}\left(4\sin^2\left(\frac {k\pi}{2^{2000}} \right)-3\right)=-1$$?
It seems so. Given this and this identities:$$\color{red}{\prod\limits_{k=1}^{2^{1999}}\left(4\sin^2\left(\frac {k\pi}{2^{2000}} \right)-3\right)}= \prod\limits_{k=1}^{2^{1999}} \left(\left(2-2\cos{\frac{k\pi}{2^{1999}}} \right)-3\right)=\\ \prod\limits_{k=1}^{2^{1999}} \left(-1-2\cos{\frac{k\pi}{2^{1999}}} \right)= \prod\limits_{k=1}^{2^{1999}} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right)=\\ -\prod\limits_{k=1}^{2^{1999}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right)=\\ -\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right) \cdot 1 \cdot \prod\limits_{k=2^{1998}+1}^{2^{1999}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right)=\\ -\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right) \cdot\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\frac{\left(2^{1998}+k\right)\pi}{2^{1999}}} \right)=\\ -\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right) \cdot\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\left(\frac{\pi}{2}+\frac{k\pi}{2^{1999}}\right)} \right)=\\ -\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right) \cdot\prod\limits_{k=1}^{2^{1998}-1} \left(1-2\cos{\left(\frac{\pi}{2}-\frac{k\pi}{2^{1999}}\right)} \right)=\\ -\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right) \cdot\prod\limits_{k=1}^{2^{1998}-1} \left(1-2\cos{\frac{\left(2^{1998}-k\right)\pi}{2^{1999}}} \right)=\\ -\prod\limits_{k=1}^{2^{1998}-1} \left(1+2\cos{\frac{k\pi}{2^{1999}}} \right) \cdot\prod\limits_{k=1}^{2^{1998}-1} \left(1-2\cos{\frac{k\pi}{2^{1999}}} \right)=\\ -\prod\limits_{k=1}^{2^{1998}-1} \left(1-4\cos^2{\frac{k\pi}{2^{1999}}} \right)= -\prod\limits_{k=1}^{2^{1998}} \left(1-4\cos^2{\frac{k\pi}{2^{1999}}} \right)=\\ -\prod\limits_{k=1}^{2^{1998}} \left(1-4\left(1-\sin^2{\frac{k\pi}{2^{1999}}} \right)\right)= -\color{red}{\prod\limits_{k=1}^{2^{1998}} \left(4\sin^2{\frac{k\pi}{2^{1999}}}-3\right)}$$
By induction: $$\prod\limits_{k=1}^{2^{1999}}\left(4\sin^2\left(\frac {k\pi}{2^{2000}} \right)-3\right)= -\prod\limits_{k=1}^{2^{1998}} \left(4\sin^2{\frac{k\pi}{2^{1999}}}-3\right)=\\ (-1)^2 \prod\limits_{k=1}^{2^{1997}} \left(4\sin^2{\frac{k\pi}{2^{1998}}}-3\right)=\\ (-1)^n \prod\limits_{k=1}^{2^{1999-n}} \left(4\sin^2{\frac{k\pi}{2^{1999-n+1}}}-3\right)=\\ (-1)^{1999} \prod\limits_{k=1}^{1} \left(4\sin^2{\frac{k\pi}{2}}-3\right)=(-1)^{1999}$$
