Suppose $K$ is a $p$-adic field (finite extension of the $p$-adics), and let $n$ be any integer (independent of what $p$ is). Define $U$ to be the set of all $x$ in $K$ such that $|x| = 1$ and such that $x = y^n$ for some $y$ in $K$. I would like to show that $U$ is an open set and that as a multiplicative group $U$ has finite index in the group of elements of $K$ of norm $1$. What's the best way of seeing why this is true (assuming it is)?
I pretty much have an idea why this holds.. in the $p$-adic case you can prove the $n$th powers are of bounded index in ${\bf Z}_{p^l}$ for each $l$ and then use an inverse limiting-type argument as $l$ goes to infinity to get this for $K = {\mathbb Q_p}$, and I think an analogous argument using powers of a uniformizer in place of powers of $p$ should work for a general $K$. But I keep thinking that this should be some well-known result or something that follows quickly from a well-known result. So I thought I'd throw this out.