Let $x \in \mathbb{Z}_p$ and $\{x_n\}$ such that, $x_n \equiv x_{n+1} (mod \, p^{n+1})$; $0 \leq x_n \leq p^{n+1}-1$ ; $|x-x_n|_p \rightarrow 0, n \rightarrow \infty$.
$|x|_p=\left(\frac{1}{p}\right)^{v_p(x)}$
Show that $x \in \mathbb{Z}_p^{*}$ iff $p \nmid x_0$.
$\Rightarrow$ We know that, $x_0 \equiv x_1 (mod \, p) \Leftrightarrow p \mid x_0 - x_1 \Leftrightarrow x_0=pt + x_1, t \in \mathbb{Z} \Rightarrow p \nmid x_0 $
$\Leftarrow $ ?
Any help is appreciated.
Hint: As you've noted,
$$\mathbb{Z}_p=\left\{(x_n)\in\prod_n \mathbb{Z}/p^n\mathbb{Z}:\text{for all }n\leqslant m,\, x_n\cong x_m\mod p^{n+1}\right\}$$
Thus, an element of $\mathbb{Z}_p$ is invertible if and only if $x_n$ is invertible for all $n$. This automatically forces $x_0\not\equiv 0\mod p$. But, conversely, if $x_0\not\equiv 0\mod p$, then $(x_n,p^n)=1$ for all $n\geqslant 1$ (why? Hint: use the consistency conditions) and thus every $x_n\in (\mathbb{Z}/p^n\mathbb{Z})^\times$
EDIT: It may also be fruitful to recall that $\mathbb{Z}_p$ is
$$\{x\in\mathbb{Q}_p:v_p(x)\geqslant 0\}$$
Since $v_p(x^{-1})=-v_p(x)$ for $x\in\mathbb{Q}_p^\times$, it should follow pretty quickly that $x^{-1}\in\mathbb{Z}_p$ for $x\in\mathbb{Z}_p$ if and only if $v_p(x)=0$ which says precisely...what?