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 $K = \mathbb{Q}\_p(\zeta\_p)$.



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> Is it true that for every $a \in K^\*$ there exists a power $q$ of $p$ such that $(a, \zeta\_p)\_q = 1$?
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Here $(,)\_q$ is the $q$-th power Hilbert symbol (norm-residue symbol) in $K(\zeta\_q) = \mathbb{Q}\_p(\zeta\_q)$.



Hilbert Symbols, Norms, and p-adic roots of unity
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. 


