Is it immediately apparent that the solution to the system of equations,

$\begin{array}{}\text{(1)}& \begin{array}{rl}{x}_{1}^{2}& ={x}_{2}+2\\ {x}_{2}^{2}& ={x}_{3}+2\\ {x}_{3}^{2}& ={x}_{4}+2\\ & \vdots \\ {x}_{n}^{2}& ={x}_{1}+2\end{array}\end{array}$

can be given by the roots of unity? Specifically,

$\begin{array}{}\text{(2a)}& x=\frac{{y}_{k}^{2}+1}{{y}_{k}}\end{array}$

where the ${y}_{k}$ are,

$\begin{array}{}\text{(2b)}& \begin{array}{rl}{y}_{k}& =\mathrm{exp}{\textstyle (}\frac{2\pi ik}{{2}^{n}-1}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=0\dots {2}^{n-1}-1\\ {y}_{k}& =\mathrm{exp}{\textstyle (}\frac{2\pi ik}{{2}^{n}+1}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=1\dots {2}^{n-1}\end{array}\end{array}$

Example. Let $n=4$. Then $(1)$ is equivalent to,

$\begin{array}{}\text{(3)}& x=((({x}^{2}-2{)}^{2}-2{)}^{2}-2{)}^{2}-2\end{array}$

Expanded out, $(3)$ is a ${2}^{4}=16$-deg polynomial and its 16 roots are given by $(2)$ where,

${y}_{k}=\mathrm{exp}{\textstyle (}\frac{2\pi ik}{15}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=0\dots 7$

${y}_{k}=\mathrm{exp}{\textstyle (}\frac{2\pi ik}{17}{\textstyle )},\phantom{\rule{thickmathspace}{0ex}}k=1\dots 8$

Ramanujan considered the system $(1)$ for $n=3,4$ in the general case and also as nested radicals. For $x=((({x}^{2}-a{)}^{2}-a{)}^{2}-a{)}^{2}-a$, see this related post. (Interestingly, $n=5$ in the general case is no longer completely solvable in radicals.)

Question:

I observed $(2)$ empirically. How do we prove from first principles that this is indeed the solution?