Finding the group of $\mathbb{Q}$-automorphisms of the field $\mathbb{Q}[x]/(x^{p-1}+x^{p-2}+\cdots +x+1)$

Question

I am trying to prove that $K:=\mathbb{Q}[x]/(x^{p-1}+x^{p-2}+\cdots+x+1)$ (with $p$ a prime) is a field and finding the $\mathbb{Q}$-automorphisms of $K$. I have some statements that I would like to see if they are correct.

1. Let $p(x)\in F[x]$ with $F$ a field. If $p(x+1)$ is irreducible on $F$ then $p(x)$ is irreducible on $F$. Proof: If $p(x)$ is reducible on $F$ then $p(x)=a(x)b(x)$ with $a(x),b(x)$ not unit. Then $p(x+1)=a(x+1)b(x+1)$ with $a(x+1),b(x+1)$ not unit. Therefore $p(x+1)$ is reducible. This is correct?

2. $K$ is a field. The polynomial $p(x):=x^{p-1}+x^{p-2}+\cdots+x+1$ is irreducible because $p(x+1)$ is irreducible by Eisenstein.

3. Because $p(x)$ is irreducible on $\mathbb{Q}$ and $\mathbb{Q}$ is a field, then $(p(x))$ is maximal. Therefore $K$ is a field.

4. Now, $\xi_{p}:=e^{(2\pi i)/p}$ is a root of $p(x)$. Therefore $K$ is isomorphic to $\mathbb{Q}(\xi_{p})$ (because the homomorphism evaluation $\varphi:\mathbb{Q}[x]\to \mathbb{Q}(\xi_{p})$ implies $\mathbb{Q}[x]/(\ker(\varphi))\simeq \mathbb{Q}(\xi_{p})$ and $\ker(\phi)=(p(x))$.

5. Because $p(x)$ is irreducible and monic, $p(x)=\min_{\mathbb{Q}}(\xi_{p})$ (minimal polynomial).

6. By 5) $[\mathbb{Q}(\xi_{p}):\mathbb{Q}]=p-1$.

And I'm stuck to find automorphisms. I know that the number of $\mathbb{Q}$-automorphism of $K$ is equal to the number of roots of the polynomial $p(x)$ (according to me there are $p-1$ roots of $p(x)$ for the fundamental theorem of algebra but I'm not sure).

Then will there be $p-1$ automorphisms?

How can I find the group of automorphisms? I understand that these automorphisms consist of permuting the different roots.

Actualization 1. I have already shown it, I attach the solution. Of course, I have a small doubt with a statement that I use.

7. The polynomial $p(x)$ has roots $\xi_{p}$, ${\xi}_{p}^2,\ldots, {\xi_{p}}^{p-1}$ in $\mathbb{Q}(\xi_{p})$. Therefore there $p-1$ $\mathbb{Q}$-automorphism of $\mathbb{Q}(\xi_{p})$. This are: $\psi_{k}:\xi_{p}\mapsto \xi_{p}^{k},\, k=1,2,\ldots, p-1$.

8. Now, $Aut(\mathbb{Q}(\xi_{p}/\mathbb{Q})=\left\{\psi_{k}:k=1,\ldots, p-1\right\}$. Let $\phi:Aut(\mathbb{Q}(\xi_{p}/\mathbb{Q})\to (\mathbb{Z}_{p})^{\times}$ with $\phi(\psi_{k})=k$. The maps $\phi$ is a isomorphism.

The question that I have is the following: Afirmation: Let $F$ field. If $f(x+1)\in F[x]$ is irreducible then $f(x)\in F[x]$ is irreducible.

My proof: If $f(x)$ is reducible then $f(x)=a(x)b(x)$ with $a(x),b(x)$ not units, then $f(x+1)=a(x+1)b(x+1)$ therefore $f(x+1)$ reducible a contradiction. This is correct? thanks