Finding the $\mathbb{Q}$-automorphisms of the splitting field of $x^p-2$ over $\mathbb{Q}$.

I have this:

The splitting field of $x^p-2$ over $\mathbb{Q}$ is $\mathbb{Q}(\sqrt[p]{2},\xi)$ with $\xi=e^{{2\pi i}/p}$.
$[\mathbb{Q}(\sqrt[p]{2})(\xi):\mathbb{Q}(\sqrt[p]{2})][\mathbb{Q}(\sqrt[p]{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[p]{2})(\xi):\mathbb{Q}(\sqrt[p]{2})]\cdot p$
$\xi$ is a root of $x^{p}-1$. Then $min_{\mathbb{Q}(\sqrt[p]{2})}(e^{{2\pi i}/p})$ divides $x^p-1$.
$x^p-1=(x-1)(x^{p-1}+x^{p-2}+\cdots +x+1)$ and $\xi$ is root of $(x^{p-1}+x^{p-2}+\cdots +x+1)$ therefore $min_{\mathbb{Q}(\sqrt[p]{2})}(e^{{2\pi i}/p})$ divides $(x^{p-1}+x^{p-2}+\cdots +x+1)$

According to me, it is true that $min_{\mathbb{Q}(\sqrt[p]{2})}(e^{{2\pi i}/p})=(x^{p-1}+x^{p-2}+\cdots +x+1)$ but i dont'know proves it (on another topic I saw that $min_{\mathbb{Q}}(e^{{2\pi i}/p})=(x^{p-1}+x^{p-2}+\cdots +x+1)$ (over $\mathbb{Q})$ but now I need to prove that it is irreducible over $\mathbb{Q}(\sqrt[p]{2})$

Let $\psi:\mathbb{Q}(\sqrt[p]{2},\xi)\to \mathbb{Q}(\sqrt[p]{2},\xi)$ automorphism such that for all $x\in \mathbb{Q},\, \psi(x)=x$.

I need find the extension of $\mathbb{Q}(\sqrt[p]{2})\to \mathbb{Q}(\sqrt[p]{2},\xi)$. Since $x^p-2=0$ implies (on $\mathbb{Q}(\sqrt[p]{2},\xi)$) that $x\in \left\{\sqrt[p]{2},\sqrt[p]{2}\xi,\sqrt[p]{2}\xi^2,\ldots, \sqrt[p]{2}\xi^{p-1}\right\}$.

By a theorem, exists $p$ extensions (because exists $p$ rootS of $x^p-2=0$) This $p$ extensions are: $\psi_1:\sqrt[p]{2}\mapsto \sqrt[p]{2},\psi_2:\sqrt[p]{2}\mapsto \sqrt[p]{2}\xi,\ldots, \psi_{p}:\sqrt[p]{2}\mapsto \sqrt[p]{2}\xi^{p-1}$

Now, I need find the extension of $\mathbb{Q}(\sqrt[p]{2})(\xi)\to \mathbb{Q}(\sqrt[p]{2},\xi)$. In this case, because $x^{p-1}+x^{p-2}+\cdots +x+1=\min_{\mathbb{Q}(\sqrt[p]{2})}(\xi)$, i have that $x^{p-1}+x^{p-2}+\cdots+x+1=0$ implies $x\in \left\{\xi,\xi^2,\ldots, \xi^{p-1}\right\}$ (therefore, exists $p-1$ extensions) The extensions are: $\bar{\psi}_{1}:\xi\mapsto \xi,\ldots, \bar{\psi}_{p-1}:\xi\mapsto \xi^{p-1}$
By 6) and 7), there are $p(p-1)$ $\mathbb{Q}$-automorphism of $\mathbb{Q}(\sqrt[p]{2},\xi)$. This are:

$\psi_{i,j}:\sqrt[p]{2}\mapsto \sqrt[p]{2}\xi^{i},\, \xi\to \xi^{j}$ where $i=0,1,2,\ldots, p-1$ and $j=1,2,\ldots, p-1.$

I would need to prove that $\min_{\mathbb{Q}(\sqrt[p]{2})}(e^{{2\pi i}/p})=x^{p-1}+x^{p-2}+\cdots +x+1$ and explicitly find the group of automorphisms (that is, see some group isomorphic to the group of automorphisms found in part 8) How can I find the group?

actualization 1. Is $x^{p-1}+x^{p-2}+\cdots+ x+1$ irreducible over irrationals (or in real numbers)? (and, by consecuence, irreducibl over $\mathbb{Q}[\sqrt[p]{2})$

$[\Bbb{Q}(\zeta_p):\Bbb{Q}]=p-1$ and $[\Bbb{Q}(2^{1/p}):\Bbb{Q}]=p$,
$p(p-1)=lcm(p,p-1)$ divides $[\Bbb{Q}(\zeta_p,2^{1/p}):\Bbb{Q}]$.
Since $\Bbb{Q}(\zeta_p,2^{1/p})$ is a quotient algebra of $\Bbb{Q}[x,y]/(x^p-2,\sum_{n=0}^{p-1} y^n)$ we know that $[\Bbb{Q}(\zeta_p,2^{1/p}):\Bbb{Q}]\le p(p-1)$.