Consider the $p$-adic field $K \supset \mathbb{Q}_p$ with ring of integers $\mathcal{O}_K$ and maximal ideal $\mathfrak{m}_K$ whose $n^{th}$ higher unit group is defined by $$ U^{(n)}=\{u \in \mathcal{O}_K^{*}~|~ u \equiv 1 \ (\text{mod} \ \mathfrak{m}_K^n \}=1+\mathfrak{m}_K^n \subset \mathcal{O}_K^{*},$$ where $n \geq 1$. The group $U^{(1)}$ is the group of principal units and $U^{(0)}:=\mathcal{O}_K^{*}$. Then we have the tower, $$ \mathcal{O}_K^{*} \supseteq U^{(1)} \supseteq U^{(2)} \supseteq \cdots $$ Consider the following quotients $$ \mathcal{O}_K^{*} /U^{(n)} \cong (\mathcal{O}_K/\mathfrak{m}_K^n)^{*} \ \text{and} \ U^{(n)}/U^{(n+1)} \cong \mathcal{O}_K/\mathfrak{m}_K.$$
Let $\bar K$ be the algebraic closure of $K$ with maximal ideal $\bar{\mathfrak{m}}_K$ be its maximal ideal.
Both the quotients $\mathcal{O}_K^{*} /U^{(n)}$ and $U^{(n)}/U^{(n+1)}$ has torsion elements and hence $\mathcal{O}_K^{*} /U^{(n)} \cap \mathcal{O}_K=\emptyset$ as well as $U^{(n)}/U^{(n+1)} \cap \mathcal{O}_K=\emptyset$.
My question:
Is $\mathcal{O}_K^{*} /U^{(n)} \cap \bar{\mathfrak{m}}_K \neq \emptyset $ ?
Thanks