Let $p$ be an odd prime number, let $\mathbb{Q}_p$ be the field of $p$-adic numbers, and let $\overline{\mathbb{Q}_p}$ be an algebraic closure of it. For a primitive $p$-th root of unity $\zeta_p \in \overline{\mathbb{Q}_p}$, set $K = \mathbb{Q}_p(\zeta_p)$.
Is it true that for every $a \in K^*$ there exists a power $q$ of $p$ such that $(a, \zeta_p)_q = 1$?
Here $(,)_q$ is the $q$-th power Hilbert symbol (norm-residue symbol) in $K(\zeta_q) = \mathbb{Q}_p(\zeta_q)$.
I think I can construct an explicit counterexample with $a\in\mathbb{Q}_p$.
Choose a compatible sequence $\zeta_{p^m}$ of $p^m$th roots of unity in $\overline{\mathbb{Q}}_p$. Write $q=p^n$ with $n\geq1$. By local class field theory, $(a,\zeta_p)_q=1$ if and only if $a$, considered as an element of $\mathbb{Q}_p(\zeta_{p^n})$, is a norm from the extension of $\mathbb{Q}_p(\zeta_{p^n})$ obtained by adjoining a $p^n$th root of $\zeta_p$. In particular, if $(a,\zeta_p)_{p^n}=1$ then $a$ is in the image of the norm map from $\mathbb{Q}_p(\zeta_{p^{n+1}})$ to $\mathbb{Q}_p(\zeta_{p^n})$, and now taking norms all the way down to $\mathbb{Q}_p$ (here is where we assume $a\in\mathbb{Q}_p$) we deduce that $a^{p^{n-1}(p-1)}$ is in the image of the norm map from $\mathbb{Q}_p(\zeta_{p^{n+1}})$ down to $\mathbb{Q}_p$.
So it suffices to find $a\in\mathbb{Q}_p$ such that for all $n\geq1$ we have that $a^{p^{n-1}(p-1)}$ is not in the image of the norm map from $\mathbb{Q}_p(\zeta_{p^{n+1}})$ down to $\mathbb{Q}_p$. Now choose $a\in\mathbb{Z}_p^\times$ a topological generator (such an element exists as $p$ is odd). Then for all $n\geq1$ we can write $a^{p^{n-1}(p-1)}=1+p^nu$ with $u\in\mathbb{Z}_p$ a unit. However (by the standard relationship between higher ramification groups, conductors and the filtration on local units) the units in $\mathbb{Q}_p$ which are norms from $\mathbb{Q}_p(\zeta_{p^{n+1}})$ are precisely those of the form $1+p^{n+1}v$ for $v\in\mathbb{Z}_p$ and in particular cannot include $a^{p^{n-1}(p-1)}$.
One does not even really care about the tame part of the story, so the above argument (if it's right) seems to show that $a=1+p$ is an explicit counterexample.
