Let $\mathbb{Q}_p$ the $p$-adic completion of $\mathbb{Q}$ and $$S=\{x\in\mathbb{Q}_p:1+x\mbox{ has $n^{th}$ roots in }\mathbb{Q}_p\mbox{ for infinitely many }n\in\mathbb{N}\}$$ I have to show that $p\mathbb{Z}_p\subseteq S\subseteq \mathbb{Z}_p$ where $\mathbb{Z}_p=\{x\in\mathbb{Q}_p:|x|_p\le 1\}$.
Any hint on how to start? I'm totally stuck at the moment, trying to find out the structure of those $n^{th}$ roots but going nowhere.
So let's start the easy way, if $x\not\in\Bbb Z_p$ we have $v_p(x)<0$, so that $v_p(1+x)<0$ by the strong triangle inequality. Then say $y_n^n=1+x$ for some infinite sequence, $y_n$. Then as
$$v_p(y_n)={1\over n}v_p(1+x)$$
we see that as $n\to\infty$ we have that $v_p(y_n)\to 0$. But the valuation is discrete, hence for large enough $n$ we must have $v_p(y)=0$ contradicting $v_p(1+x)<0$.
Now let's show that $1+pz\in S$ for every $z\in\Bbb Z_p$. Then we need only show that
$$t^n-1-pz\in\Bbb Z_p[t]$$
has a solution in $\Bbb Z_p$ for infinitely many $n$. However, we can do this by using Hensel's lemma to reduce the problem to one modulo $p$ (or $8$ in the case $p=2$) and Dedekind's theorem on infinitely many primes in arithmetic progressions.