Consequence of Hensel's Lemma The background of the question is following: About cyclotomic extensions of $p$-adic fields

Let $n$-th root $\zeta_n$ be primitive $n$-th root and $n$ coprime to the prime $p$, then there is claimed that we have an isomorphism $$\text{Gal}(\mathbb{Q}_p[\zeta_n]/\mathbb{Q}_p) \cong \text{Gal}(\mathbb{F}_p[\zeta_n]/\mathbb{F}_p)$$ between Galois groups by Hensels Lemma. 
It's not clear to me why does Hensels lemma imply this isomorphism.
 A: $\def\Gal{\operatorname{Gal}}\def\FF{\mathbb{F}}\def\QQ{\mathbb{Q}}\def\NN{\mathbb{N}}$A basic fact of finite fields is that $F=\FF_p(\zeta_n)$ (with $p \not\mid n$) is the splitting field over $\FF_p$ of $f(x) := x^{p^d-1}-1$ where $d \in \NN$ is minimal such that $n \mid p^d-1$. Also every non-zero element of $F$ is a root of $f$, and $(F:\FF_p)=d$, and $\Gal(F/\FF_p) \cong C_d$ is generated by $\zeta \mapsto \zeta^p$.
Let $E=\QQ_p(\zeta_n)$, then clearly its residue class field is $F$. Take any non-zero $\bar\alpha \in F$, it is a root of $f(x)$, and lift it to an element $\alpha \in E$, i.e. such that $\alpha$ reduces to $\bar\alpha$ in the residue class field. Now $f'(\alpha) = (p^d-1)\alpha^{p^d-2}$ and since $v_p(p^d-1)=0$ and $v_p(\alpha)=0$ we deduce that $v_p(f'(\alpha))=0$. Therefore we can apply Hensel's lemma to deduce that there is $\beta \in E$ such that $f(\beta)=0$ and $\bar\alpha = \bar\beta$. Since there were $|F|-1=p^d-1=\deg(f)$ choices of $\bar\alpha$, we have found this many distinct roots of $f(x)$ in $E$, and we deduce that $E$ contains the splitting field over $\QQ_p$ of $f(x)$. Conversely, since $\zeta_n$ is a root of $f(x)$, we see that the splitting field contains $\zeta_n$ and hence $E$. Hence $E$ is the splittng field, and so in particular $E = \QQ_p(\zeta_{p^d-1})$.
If $\sigma \in \Gal(E/\QQ_p)$ then it restricts to $\bar\sigma \in \Gal(F/\FF_p)$. Further, such a $\sigma$ must map $\zeta_{p^d-1} \mapsto \zeta_{p^d-1}^k$ for some $k$. Putting this together, then $\bar\sigma(\zeta_{p^d-1}) = \zeta_{p^d-1}^k$, but the only elements of $\Gal(F/\FF_p)$ can have $k=p^j$. We deduce that $\Gal(E/\QQ_p)$ is cyclic and generated by the automorphism $\zeta_{p^d-1} \mapsto \zeta_{p^d-1}^p$, just like $\Gal(F/\QQ_p)$ and so they are naturally isomorphic, the isomorphism being $\sigma \mapsto \bar\sigma$.
