Let $k/\mathbb{Q}_p$ be a finite extension. Let $k^{ur}$ be its maximal unratified extension and consider $K$ its completion. Let $\varphi$ be the Frobenius morphism in Gal$(k^{ur}/k)$ and whit the same letter we denote its extension to $K$; let $q$ the cardinality of the residue field of $k$ and let $v: K^\times \to \mathbb{Z}$ be the normalized valuation. Let $\pi$ be a uniformizer of $k$ and let $\xi \in k^\times$ such that $v(\xi)=d$. Let $ F_{\xi}:=\{f \in \mathcal{O}'[[x]]|f(x)=\pi'+(\text{something else}), N_{K'/K}\pi'=\xi, f(x)\equiv x^q \mod \mathfrak{p}\}$ ($\mathfrak{p}$ is the unique maximal ideal). Let $F_f$ be the Lubin Tate formal groups over $\mathcal{O}_{k'}$. After this premise I want to prove 2 statements
Let $E/k$ be an algebraic extension and $\alpha \in \widehat{E}$ an element in the completion of $E$. Show that if $\alpha$ is algebraic over $E$, then $\alpha \in E$.
I have the hint: Let $\overline{E}$ be an algebraic closure of $E$. The element $\alpha$ can be regarded as an element in the topological closure $E'$ of $E$ in $\overline{E}$. So, compare Gal$(\overline{E}/E)$ and Gal$(\overline{E}/E')$.
Let $k_{\pi^{\infty}}=k(F_f[\mathfrak{p}^{\infty}])$. Show that the composite field $L_{\pi}:=k^{ur}\cdot k_{\pi^{\infty}}$ is independent of the choice of uniformizer $\pi$.
I have the Hint: Use pre previous point and that $f'=\theta^{\varphi}\circ f \circ \theta^{-1} \in F_{\xi '}$ for any $f \in F_{\xi}$.