Roots of the polynomial $x^{p-1}+x^{p-2}+\ldots+x+1$ when $p$ is a prime. I'm studying Galois Theory and I have some doubts about the roots of the polynomial $x^{p-1}+x^{p-2}+\ldots+x+1$ when $p$ is a prime.
Let $\zeta_n$ be an $n$th root of unity, i.e, $\zeta^n-1$, and $\zeta_n \neq 1$. Then, as $(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\ldots+x+1)$, the minimal polynomial of $\zeta_n$ is $x^{n-1}+x^{n-2}+\ldots+x+1$, because it is irreducible (am I right?).
Now, let $p$ be a prime and suppose that I want to find the automorphisms of $\mathbb{Q}(\zeta_p)$ that fixes $\mathbb{Q}$, that is, the Galois group Gal($\mathbb{Q}(\zeta_p)/\mathbb{Q})$. I know that $\zeta_p$ must be send to another root of its minimal polynomial, i.e other root of $x^{p-1}+x^{p-2}+\ldots+x+1$.
Looking in some exercises, I read that $\zeta_p$ can be sent to $\zeta_p^k$ for any $k \in \{1,\ldots,p-1 \}$. But that means that $\zeta_p^k$ is a root of $x^{p-1}+x^{p-2}+\ldots+x+1$ for any $k \in \{1,\ldots,p-1 \}$.
Is that correct? How can i prove that?

The polynomial of degree $n-1$ you have written is irreducible only when $n$ is a prime number. For example  the factorization $x^4-1 = (x-1)(x+1)(x^2+1)$ tells us the degree 3 polynomial you wrote is reducible.
Now for the second part you can use De Moivre's theorem. ALternatively rewrite the equation $x^p-1=0$ as $x^p =1$. Now if $\zeta$ satisfies the equation clearly $\zeta^k$ also satisfies.. 
