The ring of $p$-adic integer is a local ring 
The ring of $p$-adic integer is a complete local ring.

I already know the ring of $p$-adic integer is complete. Problem is to show it's a local ring. I know this answer is already answered here.
In that answer, it uses one useful fact: If $P\subsetneq R$ is a proper ideal such that $x\in R\setminus P\Rightarrow x\in R^\times$, then $P$ is a unique maximal ideal i.e. $R$ is a local ring. Then it says in the case of power series $f\notin (X)\Rightarrow f\in R^\times$ so $(X)$ is the unique maximal ideal. Applying this to may case, $x\notin (p\Bbb Z_p)$ then $x\in (\Bbb Z_p)^\times$. This is not clear to me. Could you explain?
Edit: I already know $p\Bbb Z_p$ is a maximal ideal. But don't know why it's a unique maximal ideal.
Edit: I know there is some lecture notes that shows $\Bbb Z_p$ is a local ring using DVR argument but I don't want to do that. Or is that the simplest way?
 A: Let $f$ be an arbitrary element of $\mathbb{Z}_p$. We can expand it as $a_0 + a_1p + a_2p^2 + \ldots$ with the $a_i$ integers (which we can even take in $\{0, \ldots, p-1\}$ but that is irrelevant for the answer).
When $a_0 = 0$ we have $f \in p\mathbb{Z}_p$ and there is nothing to prove.
When $a_0 = 1$ we can copy the steps in the other answer to construct an element $f^{-1}$ (as an infinite series of the above form) satisfying $ff^{-1} = 1$.
What remains is the case that $a_0 \neq 0$ but we do not necessarily know that it equals $1$ either. Now in this case there is a $b \in \mathbb{Z}$ such that $a_0b \equiv 1 \mod p$. ($b$ is of course not unique, just pick one). This uses that $\mathbb{Z}/p\mathbb{Z}$ is a field.
Now it is easy to check that the element $fb \in \mathbb{Z}_p$ is of the form $1 + b_1p + b_2 p^2 + \ldots$. So by the earlier answer we can construct an element $(fb)^{-1}$ such that $(fb)(fb)^{-1} = 1$.
The last step is then observing that $f(b(fb)^{-1}) = 1$ as well, so we can take $f^{-1} = b(fb)^{-1}$.
