Question about non-archimedean local fields: how is $(1 + \mathfrak{m}^n) \langle\pi\rangle$ finite index open subgroup?

Question

Let $K$ be a non-archimedean local field. Let $\mathfrak{m}$ be the maximal ideal of $O_K$ and $\pi$ a uniformizing parameter of $O_K$. Could someone please explain me how $(1 + \mathfrak{m}^n) \langle\pi\rangle$ is an open finite index subgroup of $K^*$? Thank you very much.

Answer 1

$$U_K/(1+\mathfrak m^n)\to K^*/(1+\mathfrak m^n)\left<\pi \right>$$ induced by the inclusion $U_K\to K^*$ is an isomorphic. $$U_K/(1+\mathfrak m^n)\cong (\mathfrak O_K/\mathfrak m^n)^*$$ is finite, since $|\mathfrak O_K/\mathfrak m^n|=|k|^n$ where $k$ is the (finite) residue class field of $K$.

As $1+\mathfrak m^n$ is open, so is $(1+\mathfrak m^n)\left<\pi \right>$.