If $p\in\mathbb{Z}$ is prime, prove that:
$1)$ If $a\in\mathbb{Q}_p^{\times}$, then $a\in\mathbb{Z}_p^{\times}\iff \exists\, b_n\in\mathbb{Q}_p^{\times}$ such that $a=b_n^n$ for all $n>0$ with $\gcd(n, p(p-1))=1$
$2)$ $\text{Aut}(\mathbb{Q}_p)=\{id_{\mathbb{Q}_p}\}$
Here's what I've done:
$1)$ $(\Leftarrow)$ suppose that $a=b_n^n$ for all such $n$. Letting $|a|_p=p^k$ and $|b_n|_p=p^{e_n}$ with $k, e_n\in\mathbb{Z}$, we have that $k=ne_n$, so $e_n\to 0$ when $n\to\infty$, which means $e_{n_0}=0$ for some $n_0$. Therefore $k=n_0e_{n_0}=n_0(0)=0$, so $|a|_p=1$, which means $a\in\mathbb{Z}_p^{\times}$.
for $(\Rightarrow)$, I still have no clue.
Now $2)$: assume $1)$ is true. Taking $a\in\mathbb{Z}_p^{\times}$ and $\phi\in\text{Aut}(\mathbb{Q}_p)$, we prove that $\phi(a)\in\mathbb{Z}_p^{\times}$. If $a\tilde{a}=1$, then $\tilde{a}\in\mathbb{Z}_p^{\times}$ so we have that $\exists\,\tilde{b}_n$ with $\tilde{a}=\tilde{b}_n^n$ for every $n \geq 0$. That way, $a\tilde{a}=1\Rightarrow\phi(a)\phi(\tilde{a})=1$ $\Rightarrow \phi(a)^{-1}=\phi(\tilde{b}_n^n)=\phi(\tilde{b}_n)^n$ for all $n\geq 0$, which means $\phi(a)^{-1}\in\mathbb{Z}_p^{\times}$, thus $\phi(a)\in\mathbb{Z}_p^{\times}$. So, basically, $\phi(\mathbb{Z}_p^{\times})=\mathbb{Z}_p^{\times}$, which looks like I'm getting somewhere. But I don't know how to conclude $\phi=id_{\mathbb{Q}_p}$.
Any tips? Thanks!
For (1), hint: Hensel's lemma.
For the claim (2), it suffices to show that any automorphism $ \phi : \mathbf Q_p \to \mathbf Q_p $ must be continuous. This follows; because the fact that $ \phi $ preserves $ p $-adic units implies that it preserves $ p $-adic valuations (and by linearity, it suffices to check continuity at $ 0 $). Now, $ \phi $ fixes the dense subset $ \mathbf Q $, so it must be the identity.