Prove that a $p$-adic integer $x$ is divisible by $p^n$ if and only if $x_n = 0$ Let the ring of $p$-adic integers be the projective limit 
$$
\mathbb{Z}_p = \varprojlim_{n\in\mathbb{Z}_{\geq 1}}(\mathbb{Z}/p^n\mathbb{Z}),
$$
and denote an element $x\in\mathbb{Z}_p$ as a sequence $x=(x_n)_{n\in\mathbb{Z}_{\geq 1}}$ with $x_n \in \mathbb{Z}/p^n\mathbb{Z}$.
I want to prove the following:

An element $x \in \mathbb{Z}_p$ is divisible by $p^n$ if and only if $x_n=0$.

The implication to the right is clear. For the implication to the left, it is easily shown that $x_i$ is divisible by $p^n$ for all $i> 0$. Further, if $x_i=p^nx_i'$ for some $x_i'\in\mathbb{Z}/p^i\mathbb{Z}$ (and there may be multiple choices for such $x_i'$, since $\mathbb{Z}/p^i\mathbb{Z}$ is not a domain) and $x'_j := (x'_i \bmod p^j)$ for $j<i$ we have that $p^j x_j' = x_j$. But there does not seem to be a way to construct a $p$-adic integer $x'$ such that $x'=p^nx$ from this: when increasing our index $i$ the choices of the previous elements $x_j'$ might change ad infinitum. With the approach I've taken so far, is there a way to finish the proof?
See Serre, A Course in Arithmetic, chapter II proposition 1, for a proof using a different approach.
 A: A $p$-adic integer is not any sequence $(x_n)_{n\ge1}$, but it has to satisfy compatibility conditions:
$$x_n\bmod p^{n-1}=x_{n-1}.$$
Hence if $x_n\equiv 0\mod p^n$, $x_n\equiv 0\mod p^{n-1}. $
A: The compatibility condition is crucial, so let’s get it on the board. You’re talking about sequences $(x_1,x_2,x_3,\cdots)$ such that: for each $m$, $x_m\in\Bbb Z/p^m\Bbb Z$ and $x_{m+1}\equiv x_m\pmod{p^m}$. Note that this condition also implies that for each $m$ and each $k\ge m$, $x_m\equiv x_k\pmod{p^m}$.
Let’s look at your problem now, in the special case that $n=2$. You have a sequence $(x_1,0,x_3,x_4,\cdots)$ satisfying compatibility. But $m=1$ says that $x_1\equiv0\pmod p$, so our sequence is actually $(0,0,x_3,x_4,\cdots)$. Furthermore, $m=2$ says that $x_3\equiv0\pmod{p^2}$, so it can be represented by $p^2\xi_3$ for $\xi_3\in\Bbb Z$, and indeed we can take $0\le\xi_3<p$. Similarly, $m=3$ says that $x_4\equiv x_2\pmod{p^2}$ and also that $x_4\equiv x_3\pmod{p^3}$. Thus $x_4=p^2\xi_4$  (where we can even take $0\le\xi_4<p^2$), and the congruence $p^2\xi_4\equiv p^2\xi_3\pmod{p^3}$ is equivalent to $\xi_3\equiv\xi_4\pmod p$. And so on.
I was going to go farther along these lines, but it seems to me that I’ve already answered your question. If not, I’ll be glad to amplify.
Amplification: I have made some changes in the text above, as an aid to the following:
The arguments above show that $\xi=(\xi_3,\xi_4,\xi_5,\cdots)$ is a good element of $\Bbb Z_p$. I claim that the original $x=(0,0,x_3,x_4,\cdots)$ is equal to $p^2\xi$. Certainly, $p^2\xi=(p^2\xi_3,p^2\xi_4,p^2\xi_5,\cdots)$ with $p^2\xi_3\equiv0\pmod p$ and $p^2\xi_4\equiv0\pmod{p^2}$. What about the other coordinates? For $m\ge3$, the $m$-th coordinate of $p^2\xi$ is $p^2\xi_{m+2}=x_{m+2}\equiv x_m\pmod{p^m}$, and I think that that does it.
