Powers in finite extensions of the p-adics Fix a prime $p$. A p-adic field is a finite extension of $\mathbb{Q}_p$.
Question 1: Let $K$ be a $p$-adic field and fix $n$. Is there $m$ such that if $\alpha \in \mathbb{Q}_p$ is an $m$th power in $K$ then $\alpha$ is an $n$th power in $\mathbb{Q}_p$? $\quad$ (Probably $m$ is a multiple of $n$.)
I think that you'd approach this by decomposing $K/\mathbb{Q}_p$ into a tower of extensions of some sort and then go up the tower. So we probably want to actually prove the following.
Question 2: Suppose that $K/L$ is an extension of $p$-adic fields and fix $n$. Is there $m$ such that if $\alpha \in K$ is an $m$th power in $L$ then $\alpha$ is an $n$th power in $K$?
I can prove Question 1 in the case when $K/\mathbb{Q}_p$ is unramified, but I don't know about the general case. I think that one might be able to use class field theory, or maybe it's already well-known, or maybe I am just missing something elementary.
I have good motivation for this question, but it would take a bit of work to explain, the motivation comes from logic.
 A: Yes.
I'll solve the more general question 2.
Let $v_p(n)$ be the highest power of $p$ dividing $n$. Then if $\alpha \in L$ is congruent to $1$ modulo $p^{v_p(n)+1}$, then $\alpha$ is an $n$th power in $L$.
Every $\alpha \in L$ can be written as $\pi^j u$ where $j\in \mathbb Z$ and $u$ is a unit in $L$. It suffices to find $m$ such that if $\pi^j u$ is an $m$th power in $K$ then $j$ is divisible by $n$ and $u$ is congruent to $1$ modulo $p^{v_p(n)+1}$.
Let $e$ be the ramification degree of $K$ over $L$, i.e. the greatest $e$ such that there is an element of $K$ whose $e$'th power is $\pi$ times a unit and let $k$ be any natural number. Then if $\pi^j u$ is a $ke$'th power in $K$ then $j$ is divisible by $k$, and hence $u = \pi^j u / (\pi^{j/k})^k$ is also a $k$th power in $K$.
So it suffices to find $k$ such that if $u$ is a unit that is a $k$th power in $K$ then $u$ is congruent to $1$ mod $p^{v_p(n)+1}$.
Let $q$ be the order of the residue field in $k$. If $u$ is a $k$th power, then it is a $k$th power of some unit $v$, and $v^{q-1}$ is congruent to $1$ modulo the uniformizer $\pi'$ of $K$. So $$ v^{ (q-1) p^r} = (1 + (v^{q-1}-1))^{p^r} = \sum_{i=0}^{P^r} \binom{p^r}{i} (v^{q-1}-1)^i$$ where the $i$th term is divisible by $\binom{p^r}{i}  \pi'^{i}$.
Taking $r$ sufficiently large, we can ensure that $\binom{p^r}{i}  \pi'^{i}$ is divisible by $p^{ v_p(n)+1}$ for all $i>0$, so $v^{ (q-1) p^r}$ is congruent to $1$ mod $p^{ v_p(n)+1}$,  so we may take $k= (q-1) p^r$.
