Open subgroups in local fields Let $K$ be a local field. How to prove that $(K^*)^n$ is open in $K$ if and only if $\mathrm{char}(K)$ does not divide $n$?
 A: I’ll give a relatively sketchy answer, depending on you to fill in the details. Accordingly, I’ll deal only with the case of characteristic $p>0$, since the mixed-characteristic case is not so interesting in this regard.
I think you will find it easy to show that $(K^*)^n$ is open in $K^*$ if and only if $(1+\mathfrak p)^n$ is open in $1+\mathfrak p$. (Here, $\mathfrak p$ is the maximal ideal of the ring of integers of $K$.) So I deal with this multiplicative subgroup of $K^*$ only.
If $n$ is prime to $p$, then $(1+\mathfrak p)^n$, the set of $n$-th powers of elements of $1+\mathfrak p$, is equal to $1+\mathfrak p$ itself: that is, every element of $1+\mathfrak p$ has an $n$-th root in $1+\mathfrak p$. You can see this with Hensel’s Lemma applied to $X^n-(1+\varepsilon)$, with $\varepsilon\in\mathfrak p$, or by noticing that the Binomial expansion for $(1+t)^{1/n}$ has no $p$ in the denominators of the coefficients.
If $p$ divides $n$, then you know that $(1+\mathfrak p)^p$ is just $1+\kappa[[\pi^p]]$, where $\kappa$ is the constant field of $K$. And there are elements of $K=\kappa((\pi))$ arbitrarily close to $1$ that are not in $1+\kappa[[\pi^p]]$, for instance $1+\pi^{Np+1}$ for $N$ as large as you like.
EDIT: My answer to your question about the equivalence between openness of $(K^*)^n$ in $K^*$ and openness of $(1+\mathfrak p)^n$ in $1+\mathfrak p$ :
First, $(K^*)^n\cap\mathscr O^*=(\mathscr O^*)^n$, for if something of valuation $0$ is an $n$-th power, it’s the $n$-th power of an element of valuation $0$. Therefore, if $(K^*)^n$ is open in $K^*$, its intersection with $\mathscr O^*$ will be open in $\mathscr O^*$, and so $(\mathscr O^*)^n$ will be open in $\mathscr O^*$.
Next, suppose that $(\mathscr O^*)^n$ is open in $\mathscr O^*$. As above, $(\mathscr O^*)^n\cap1+\mathfrak p$ is open in $1+\mathfrak p$. I now need only show that $(\mathscr O^*)^n\cap1+\mathfrak p=(1+\mathfrak p)^n$. Indeed, let $z\in\mathscr O$ with $z^n\in1+\mathfrak p$. If $q$ is the cardinality of $\mathscr O/\mathfrak p$, then $n|(q-1)$, and $z\equiv\zeta\pmod{\mathfrak p}$, where $\zeta$ is an $n$-th root of unity in $\mathscr O$. Then $z/\zeta\in1+\mathfrak p$ and $(z/\zeta)^n=z^n$, which is what I needed to show. ( That was not as easy as I expected when I wrote “I think you’ll find it easy”. )
