Maximal tamely ramified extension of $\mathbb{Q}_p$ Let $p>2$ be a prime number. Now the maximal unramified extension of $\mathbb{Q}_p$ is
$$
K^{\mathrm{unr}}=\bigcup_{(n,p)=1}\mathbb{Q}_p (\zeta_n)
$$
where $\zeta_n$ is a primitive $n$-th root of unity. We can describe the maximal tamely ramified extension of $\mathbb{Q}_p$ as
$$
K^{\mathrm{tr}}=\bigcup_{(n,p)=1}K^{\mathrm{unr}}(\sqrt[n]{p})
$$
Is it true then the only $p$-power root of unity in $K^{\mathrm{tr}}$ is $1$?
 A: The extension $\mathbf Q(\zeta_p)$ is tamely ramified ($e = p-1$).  So perhaps your question should be: are the only $p$-power roots of unity in $K^{\rm tr}$ the $p$-th roots of unity? They are.
If $\zeta$ is a $p$-power root of unity then $\mathbf Q_p(\zeta)$ is a totally ramified extension of $\mathbf Q_p$. So if this field lies in $K^{\rm tr}$, in which every finite extension of $\mathbf Q_p$ is tamely ramified, the field $\mathbf Q_p(\zeta)$ is both totally ramified and tamely ramified.  Letting $\zeta$ have order $p^n$, if $n \geq 1$ then the ramification index of $\mathbf Q_p(\zeta)/\mathbf Q_p$ is $p^{n-1}(p-1)$ and also this ramification index has to be relatively prime to $p$ (since it is tamely ramified). That implies $n = 1$, so $\zeta$ has order $p$ if $\zeta \not= 1$.  Thus the only possible $p$-power roots of unity in $K^{\rm tr}$ are the $p$-th roots of unity, and conversely the $p$-th roots of unity really are in $K^{\rm tr}$.
You assumed $p > 2$, but this argument works if $p = 2$.
