"Golden Ratio" for $n$ subsegments

According to Wikipedia, the root between 0 and 1 of $x^3+x^2+x-1=0$ is $$\bigl(\root3\of{17+3\sqrt{33}}-\root3\of{-17+3\sqrt{33}}-1\bigr)/3=0.543689012$$ to nine decimals. The root between 0 and 1 of $x^4+x^3+x^2+x-1=0$ is the reciprocal of $$p_1+(1/4)+\sqrt{(p_1+(1/4))^2-(2\lambda_1/p_1)(p_1+(1/4))+(7/(24p_1))+(1/6)}$$ where $p_1=\sqrt{\lambda_1+(11/48)}$ and $$\lambda_1={\root3\of{3\sqrt{1689}-65}-\root3\of{3\sqrt{1689}+65}\over12\root3\of2}$$ According to D. A. Wolfram, "Solving Generalized Fibonacci Recurrences", Fib. Quart. May 1998 129-145, (see in particular page 136), for $5\le n\le11$ the Galois group is the symmetric group on $n$ letters, which is not a solvable group, from which it follows that for these values of $n$ there is no solution in radicals. Wolfram conjectures the Galois group is $S_n$ for all $n\ge5$, from which it would follow that there is no solution in radicals for any $n\ge5$.

I would not be surprised to learn that in the meantime someone has proved this conjecture. If I find it, I'll get back to you. Progress is made in the paper Paulo A. Martin, The Galois group of $x^n-x^{n-1}-\cdots-x-1$, J. Pure Appl. Algebr. 190 (2004) 213-223. Martin proves the conjecture if $n$ is even, and if $n$ is prime. The paper may be available at https://www.sciencedirect.com/science/article/pii/S0022404903002457 (or it may be behind a paywall).

2) The answer is yes. Indeed, if we assume that the initial segment has unit length and the first (the largest) subdivision segment has length $a$ then the second subdivision segment has length $a\cdot a=a^2$, the third subdivision segment has length $a^2\cdot a=a^3$, and so forth. Since the sum of lengths of the subdivision segments is $1$, we obtain the respective equality. It can be simplified to

$$\frac {1-a^{n+1}}{1-a}-2=0$$

$$a^{n+1}-2a+1=0, 0<a< 1.$$

I guess for (almost) all $n>4$ this equation has no closed form solution in radicals, but I don’t know Galois theory sufficiently well to show this. Another approach to look for a closed form is to put $a=\cos t$, but I don’t see how we can proceed farther from this.