Let $p$ be an odd prime and let $n\geq 1$. Set $K=\mathbb{Q}_p(\zeta_{p^n})$, $L=\mathbb{Q}_p(\sqrt[p^n]{p})$, and $M=KL$. I claim that $M$ is totally ramified of degree $\phi(p^n)p^n$ (the proof simultaneously shows that $K$ and $L$ are linearly disjoint). Assume not, then there exists an unramified extension $W\subseteq M$ with $[W:\mathbb{Q}_p]=f>1$ ($f$ being necessarily a power of $p$). By the structure theorem for unramified extensions of $\mathbb{Q}_p$, we know that $W=\mathbb{Q}_p(\zeta_{r})$ where $r=p^f-1$. The extension $K$ is totally ramified over $\mathbb{Q}_p$ and therefore linearly disjoint from $W$. We thus have $$ \mathbb{Q}_p\subseteq K \subsetneqq KW \subseteq M=K(\sqrt[p^n]{p}) $$ By Galois theory, since $f>1$, it follows (this also holds true if $K$ and $L$ are not linearly disjoint) that $\sqrt[p]{p}\in KW$ and therefore $\mathbb{Q}_p(\zeta_p,\sqrt[p]{p})\subseteq KW$. But $KW$ is abelian over $\mathbb{Q}_p$ and $\mathbb{Q}_p(\zeta_p,\sqrt[p]{p})$ is not. Contradiction.
Q1: How to construct (systematically in $p$ and $n$) an explicit uniformizer of $M$ (for $n>1$ of course)?
Q2 Is it possible to find some uniformizer $\pi'$ of $K$ (so $v_p(\pi')=\frac{1}{\phi(p^n)}$) such that $K(\sqrt[p^n]{\pi'})=M$ (for $n>1$ of course) ?
Remark: Note that Q2 is equivalent to find (by Lagrange's resolvent) $\beta\in M$ such that $$ \sum_{j=0}^{p^n-1} \sigma^j(\beta)\zeta_{p^n}^j, $$ is a uniformizer of $M$. Here $\sigma$ is a generator of $Gal(M/K)$. This seems to boil down to some difficult linear algebra.
