Set of integer p-adics-Proposition 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."
Proof:
We will show that $\bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$. 
If $x=(\overline{x_k}) \in p^n \mathbb{Z}_p$, then it is $0=x_k \in \mathbb{Z}/p^{k+1}\mathbb{Z}$ for $k<n$. Therefore it holds for a $x$ $x_k=0$ for all $k \in \mathbb{N}_0$, thus $x=0$. 
Now let any ideal $I \neq 0$. From $\bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=0$ it follows that there is a $n \in \mathbb{N}_0$ with $I \subseteq p^n \mathbb{Z}_p$, but $I \nsubseteq p^{n+1}\mathbb{Z}_p$. 
We claim that $I=p^n \mathbb{Z}_p$. From $I \nsubseteq p^{n+1}\mathbb{Z}_p$ follows the existence of a $x=p^nu \in I$ with $u \in \mathbb{Z}_p^{\star}$. Thus it holds $p^n \in I$ and so $p^n \mathbb{Z}_p \subseteq I$. 
Since the other inclusion depends on the choice of $n$, we have $I=p^n \mathbb{Z}_p$. 
Now we consider the canonical projection 
$$\pi: \mathbb{Z}_p \subseteq \Pi_{k \in \mathbb{N}_0} \mathbb{Z}/p^{k+1}\mathbb{Z} \rightarrow \mathbb{Z}/ p^n \mathbb{Z}$$ 
It is obviously surjective since $\overline{x_{n-1}} \in \mathbb{Z}/p^n \mathbb{Z}$ is the image of $x_{n-1} \in \mathbb{Z}_p$. The kernel of $\pi$ is an ideal and thus of the form $ker \pi=p^l \mathbb{Z}_p$. Since $\pi(p^k)=0$ exactly then when $k \geq n$, it holds $ker \pi=p^n \mathbb{Z}_p$. From the fundamental theorem on homomorphisms we get that $\mathbb{Z}_p/ p^n \mathbb{Z}_p \cong \mathbb{Z}/p^n \mathbb{Z}$. 
Could you maybe explain to me the above proof?
 A: Let's look at the definition of $\mathbb{Z}_p:$
$$
\mathbb{Z}_p := \{(x_n)_{n \geq 0} \in \Pi_{n \in \mathbb{N}_0}\mathbb{Z}/p^{n+1}\mathbb{Z} | x_n = \eta_{mn}(x_m), \forall m \geq n  \}
$$
where for $m \geq n, \eta_{mn}: \mathbb{Z}/p^{m}\mathbb{Z} \rightarrow \mathbb{Z}/p^{n}\mathbb{Z}$ is the natural map induced by the inclusion $p^m\mathbb{Z} \subseteq p^n \mathbb{Z}.$
Let $x = (x_k)_{k \geq 0} \in \bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p.$ Fix one $n \in \mathbb{N}_0.$ Now $x \in p^n\mathbb{Z}_p \Rightarrow x_1 = 0, x_2 =0, \cdots , x_{n} = 0$ ($x_k \in \mathbb{Z}/p^{k+1}\mathbb{Z} \Rightarrow p^nx_k = 0$ for $k < n$). Now this is true for every $n.$ So $x_k = 0, \forall k \Rightarrow x = 0.$
Note that every element of $\mathbb{Z}_p$ can be written uniquely of the form $b_0 + b_1p + b_2p^2 + b_3p^3 + \cdots$ with each $b_i \in \{0, 1, 2, \cdots , p-1 \}.$ From this it follows that an element of $\mathbb{Z}_p$ is unit if and only if $b_0 \neq 0$ (here we need $\mathbb{Z}_p$ is an integral domain). Let $I$ be a non-zero proper ideal of $\mathbb{Z}_p.$ Then $I \subseteq p\mathbb{Z}_p.$ Choose maximal $k \in \mathbb{N}$ such that $p^k$ divides all the elements of $I.$ Such a $k$ exists because $I \neq (0), I \subseteq p\mathbb{Z}_p$ and $\bigcap_{n \in \mathbb{N}_0} p^n \mathbb{Z}_p=(0).$ Then $I \subseteq p^k\mathbb{Z}_p$ and $I \nsubseteq p^{k+1}\mathbb{Z}_p.$ Choose an element $x \in p^k\mathbb{Z}_p$ such that $x \notin p^{k+1}\mathbb{Z}_p.$ Then $x = p^ky$ where $y$ is a unit in $\mathbb{Z}/p\mathbb{Z}.$ So $p^k = xy^{-1} \in I \Rightarrow p^k\mathbb{Z}_p \subseteq I.$
Consider the map $\pi : \mathbb{Z}_p \rightarrow \mathbb{Z}/p^k\mathbb{Z}$ induced by restricting the natural surjection $\Pi_{n \in \mathbb{N}_0} \mathbb{Z}/p^n \mathbb{Z} \rightarrow \mathbb{Z}/p^k\mathbb{Z}.$ This map is clearly a surjection. We want to find the kernel of this map. First note that $p^k\mathbb{Z}_p \subseteq$ ker $\pi.$ Since ker $\pi$ is an ideal of $\mathbb{Z}_p$ and $\pi (p^{k-1}\mathbb{Z}_p) \neq (0),$ using the classification of ideals of $\mathbb{Z}_p$ we get that ker $\pi = p^k\mathbb{Z}_p.$
