Minimal Polynomial of $1+2^{\frac{1}{n}} +\cdots+2^{\frac{n-1}{n}}$

Question

I want to find the minimal polynomial of $1+2^{\frac{1}{n}} +\cdots+2^{\frac{n-1}{n}}$ over $\mathbb {Q}$. So I used Wolfram Alpha to observe that it is $$x^n -\sum \limits _{i=1}^{n} {n \choose i} x^{n-i}=0$$ But I am finding it difficult to prove this. Can anyone give any hints. Thanks.

Edit $ 1$: I have feeling that all the conjugate roots are $1+ {\zeta}2^{\frac{1}{n}} +\cdots+{\zeta}^{\frac{n-1}{n}}2^{\frac{n-1}{n}}$. And the rotations of only the $\zeta$ keeping the $1$ fixed that is one more root I guess would be $1+ {\zeta}^22^{\frac{1}{n}} + {\zeta}^32^{\frac{2}{n}} +\cdots+{\zeta}2^{\frac{n-1}{n}}$. If we assume this then the sum of roots is $n$. But I do not know whether this is true.

Answer 1

$$\alpha=1+2^{\frac{1}{n}} +....+2^{\frac{n-1}{n}}=\frac{1}{2^{\frac{1}{n}}-1}$$ From that we deduce $$2\alpha^n-(1+\alpha)^n=0$$ and this shows $\alpha$ is a root of the polynomial $x^n -\sum \limits _{i=1}^{n} {n \choose i} x^{n-i}=0$

Answer 2

$\displaystyle\sum_{k=0}^{n-1}a^k=\frac{a^n-1}{a-1}.~$ For $a=\sqrt[\large n]2,~$ this becomes $~\dfrac1{\sqrt[\large n]2-1}.~$ Now, what is the minimal polynomial

of $\sqrt[\large n]2-1,~$ and what can we infer from that ? :-$)$