As we already know, the following polynomial is irreducible over $\mathbb Q[x]$:
$$x^n + x^{n-1} + \cdots + x^2 + x - 1 = \frac{x^{n+1} - 2x + 1 }{ x-1}.$$
By Descartes' rule of signs, it has only 1 positive real root. It seems, though I do not know how to prove it yet, that there is only 1 real root if $n$ is odd and 2 real roots of opposite signs if $n$ is even.
Computational tests, conducted with GAP, suggest that the Galois group of this polynomial is $S_n$. How we can actually prove it?
Any help is greatly appreciated.