= Solution
For odd $p$ there is a decomposition
$$
\mathbb Q_p^\times
\cong p^{\mathbb Z}\times\mu_{p-1}\times(1+p\mathbb Z_p)
$$
into the valuation factor, the <Teichmuller representative> factor, and the group of <principal units>. If $x$ is a $((p-1)m)$th power for every $m$ coprime to $p$, its valuation is divisible by every such $(p-1)m$, and is therefore zero. Its residue in $\mathbb F_p^\times$ is a $(p-1)$st power, hence is one. Thus $x\in1+p\mathbb Z_p$.
Conversely, exponentiation by any integer $n$ coprime to $p$ is an automorphism of $1+p\mathbb Z_p$. This follows either from the <principal-unit logarithm>, under which it becomes multiplication by $n$, or by applying the Hensel lemma to $Y^n-x$. Since $(p-1)m$ is coprime to $p$, every $x\in1+p\mathbb Z_p$ has a $((p-1)m)$th root for every allowed $m$.
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