@DanielFischer An integral domain is called principal ideal domain , if each ideal $I\subseteq A$ is a principal ideal, i.e. there is a $x\in A$, such that $I=A\cdot x=\left\{a\cdot x\mid a\in A\right\}$.
In our case it can be $I=0$ or $I=p^n \mathbb{Z}_p$.
If $I=0$, we can write $I=\mathbb{Z}_p \cdot 0=\left\{a\cdot 0\mid a\in \mathbb{Z}_p\right\}$, right?
If $I=p^n \mathbb{Z}_p$, we can write $I=\mathbb{Z}_p\cdot p^n=\left\{a\cdot p^n\mid a\in \mathbb{Z}_p\right\}$.
So, $\mathbb{Z}_p$ is a PID.. Or am I wrong?

This is an other proposition of the theorem I am looking at... I will send you the proof.. But if it would be known, could we justify like that the fact that $\mathbb{Z}_p$ is a PID?

This is the proof of the proposition "$\mathbb{Z}_p$ contains only the ideals $0$ and $p^n \mathbb{Z}_p$ for $n \in \mathbb{N}_0$. It holds $\bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ and $\mathbb{Z}_p \ p^n\mathbb{Z}_p \cong \mathbb{Z} \ p^n \mathbb{Z}$.
Especially $p\mathbb{Z}_p$ is the only maximal ideal.":

@DanielFischer Do you mean like that? :/

An integral domain is called principal ideal domain , if each ideal $I\subseteq A$ is a principal ideal, i.e. there is a $x\in A$, such that $I=A\cdot x=\left\{a\cdot x\mid a\in A\right\}$.
In our case it can be $I=0$ or $I=p^n \mathbb{Z}_p$.
If $I=0$, we can write $I=\mathbb{Z}_p \cdot 0=\left\{a\cdot 0\mid a\in \mathbb{Z}_p\right\}$, right?
If $I=p^n \mathbb{Z}_p$, we can write $I=\mathbb{Z}_p\cdot p^n=\left\{a\cdot p^n\mid a\in \mathbb{Z}_p\right\}$.

@DanielFischer We have the canonical function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p, x \mapsto (\overline{x})_{k \in \mathbb{N}_0}=(\overline{x}, \overline{x}, \overline{x}, \dots )$.
The function $\epsilon_p: \mathbb{Z} \to \mathbb{Z}_p$ is an embedding. We interpret with it $\mathbb{Z}$ as a subset of $\mathbb{Z}_p$.

Proof:
Let $x \in \mathbb{Z}$ with $\epsilon_p(x)=(\overline{x})_k=0$, i.e. $x \equiv 0 \mod{p^{k+1}}$ for all $k \in \mathbb{N}_0$. The only integer number that is divisible by any high $p$-power is zero, so it holds $x=0$. Thus, $\epsilon_p$ is an embedding.