[proofplan] We prove that $\mathbb{Z}_p$ is $p$-adically complete and is isomorphic as a topological ring to the inverse limit $\varprojlim \mathbb{Z}/p^n\mathbb{Z}$. The proof proceeds in three stages. First, we construct a canonical map $\nu: \mathbb{Z}_p \to \varprojlim \mathbb{Z}_p / p^n \mathbb{Z}_p$ and show it is injective (an element mapping to zero in every quotient must have arbitrarily small absolute value, hence equals zero). Second, we prove surjectivity by constructing a convergent $p$-adic series from a coherent sequence of residues. Third, we invoke the previously established isomorphisms $\mathbb{Z}/p^n\mathbb{Z} \cong \mathbb{Z}_p/p^n\mathbb{Z}_p$ to pass from the inverse limit of $\mathbb{Z}_p$-quotients to the inverse limit of $\mathbb{Z}$-quotients. [/proofplan] [step:Define the canonical map $\nu$ and prove injectivity] The quotient maps $q_n: \mathbb{Z}_p \to \mathbb{Z}_p / p^n \mathbb{Z}_p$ for $n \geq 1$ are compatible with the transition maps $\mathbb{Z}_p / p^{n+1}\mathbb{Z}_p \to \mathbb{Z}_p / p^n \mathbb{Z}_p$ (induced by $p^n \mathbb{Z}_p \supseteq p^{n+1}\mathbb{Z}_p$). By the universal property of the inverse limit, there is a unique ring homomorphism \begin{align*} \nu: \mathbb{Z}_p &\to \varprojlim_{n} \mathbb{Z}_p / p^n \mathbb{Z}_p \end{align*} with $\pi_n \circ \nu = q_n$ for all $n \geq 1$, where $\pi_n$ denotes the $n$-th projection from the inverse limit. Explicitly, $\nu(x) = (x + p^n \mathbb{Z}_p)_{n \geq 1}$. To show $\nu$ is injective, let $x \in \ker(\nu)$. Then $q_n(x) = 0$ for all $n \geq 1$, meaning $x \in p^n \mathbb{Z}_p$ for all $n \geq 1$. This gives $|x|_p \leq p^{-n}$ for all $n \geq 1$. Letting $n \to \infty$, we obtain $|x|_p = 0$, so $x = 0$ by the non-degeneracy of the absolute value. Therefore $\nu$ is injective. [/step] [step:Prove surjectivity by constructing a convergent $p$-adic expansion] Let $(z_n)_{n \geq 1} \in \varprojlim \mathbb{Z}_p / p^n \mathbb{Z}_p$ be a coherent sequence, meaning $z_n \in \mathbb{Z}_p / p^n \mathbb{Z}_p$ and $z_{n+1} \equiv z_n \pmod{p^n \mathbb{Z}_p}$ for all $n \geq 1$. For each $n \geq 1$, choose a representative $\tilde{z}_n \in \mathbb{Z}_p$ of the coset $z_n$. Since $\mathbb{Z}$ is dense in $\mathbb{Z}_p$ (by the theorem [$\mathbb{Z}_p$ Is the Closure of $\mathbb{Z}$ in $\mathbb{Q}_p$](/theorems/???)), we may choose $\tilde{z}_n \in \{0, 1, \ldots, p^n - 1\}$ (these integers form a complete set of representatives for $\mathbb{Z}_p / p^n \mathbb{Z}_p$, as follows from the isomorphism $\mathbb{Z}/p^n\mathbb{Z} \cong \mathbb{Z}_p / p^n \mathbb{Z}_p$). Write $x_n = \tilde{z}_n$. The compatibility condition $z_{n+1} \equiv z_n \pmod{p^n}$ ensures $x_{n+1} \equiv x_n \pmod{p^n}$, so $x_{n+1} - x_n = a_n p^n$ for some $a_n \in \{0, 1, \ldots, p-1\}$ (since $0 \leq x_{n+1} < p^{n+1}$ and $0 \leq x_n < p^n$). The digits $a_i \in \{0, 1, \ldots, p-1\}$ for $i \geq 0$ are determined by $x_1 = a_0$ and $x_{n+1} = \sum_{i=0}^{n} a_i p^i$. Define $x = \sum_{i=0}^{\infty} a_i p^i$. This series converges in $\mathbb{Z}_p$ because $|a_i p^i|_p \leq |p^i|_p = p^{-i} \to 0$ as $i \to \infty$ (convergence of series in a non-archimedean valued field requires only that the terms tend to zero, by the theorem [Convergence via Term Decay](/theorems/???)). Moreover, for each $n \geq 1$, \begin{align*} x - x_n = \sum_{i=n}^{\infty} a_i p^i, \end{align*} and $|x - x_n|_p \leq \max_{i \geq n} |a_i p^i|_p \leq p^{-n}$, so $x \equiv x_n \pmod{p^n \mathbb{Z}_p}$. Therefore $\nu(x) = (x + p^n \mathbb{Z}_p)_{n \geq 1} = (z_n)_{n \geq 1}$, and $\nu$ is surjective. [guided] We need to show that every coherent sequence $(z_n)$ in the inverse limit has a preimage under $\nu$. The idea is to build an element of $\mathbb{Z}_p$ digit by digit, using the compatibility of the coherent sequence. The isomorphism $\mathbb{Z}/p^n\mathbb{Z} \cong \mathbb{Z}_p/p^n\mathbb{Z}_p$ tells us that the integers $\{0, 1, \ldots, p^n - 1\}$ form a complete set of coset representatives for $\mathbb{Z}_p / p^n \mathbb{Z}_p$. So for each $n$, there is a unique integer $x_n \in \{0, 1, \ldots, p^n - 1\}$ representing $z_n$. The coherence condition $z_{n+1} \equiv z_n \pmod{p^n}$ translates to $x_{n+1} \equiv x_n \pmod{p^n}$. Since $0 \leq x_n < p^n$ and $0 \leq x_{n+1} < p^{n+1}$, we can write $x_{n+1} = x_n + a_n p^n$ for a unique $a_n \in \{0, 1, \ldots, p-1\}$. Unrolling: $x_n = a_0 + a_1 p + a_2 p^2 + \cdots + a_{n-1} p^{n-1}$, which is the base-$p$ expansion of $x_n$. The formal power series $x = \sum_{i=0}^{\infty} a_i p^i$ converges in $\mathbb{Z}_p$. Why does this series converge? By the [Convergence via Term Decay](/theorems/???) theorem, a series in a complete non-archimedean field converges if and only if its terms tend to zero. Here $|a_i p^i|_p = |a_i|_p \cdot p^{-i} \leq p^{-i} \to 0$ (since $a_i \in \{0, \ldots, p-1\} \subset \mathbb{Z}$ gives $|a_i|_p \leq 1$). So the series converges. The partial sums of this series are exactly the $x_n$: \begin{align*} \sum_{i=0}^{n-1} a_i p^i = x_n. \end{align*} The tail $x - x_n = \sum_{i=n}^{\infty} a_i p^i$ satisfies $|x - x_n|_p \leq \max_{i \geq n} p^{-i} = p^{-n}$, so $x \equiv x_n \pmod{p^n \mathbb{Z}_p}$. This means $\nu(x) = (z_n)$, completing the surjectivity argument. [/guided] [/step] [step:Verify $\nu$ is a topological isomorphism] Since $\nu$ is a bijective ring homomorphism, it remains to show it is a homeomorphism. The inverse limit topology on $\varprojlim \mathbb{Z}_p / p^n \mathbb{Z}_p$ is the subspace topology inherited from $\prod_{n \geq 1} \mathbb{Z}_p / p^n \mathbb{Z}_p$, where each factor carries the discrete topology. A basis for this topology consists of sets $\pi_n^{-1}(\{c\})$ for $c \in \mathbb{Z}_p / p^n \mathbb{Z}_p$. The preimage $\nu^{-1}(\pi_n^{-1}(\{c\})) = q_n^{-1}(\{c\})$ is a coset of $p^n \mathbb{Z}_p$ in $\mathbb{Z}_p$, which is open (and closed) in the $p$-adic topology. So $\nu$ is continuous. For the inverse: the map $\nu^{-1}$ is continuous because the basic open sets in $\mathbb{Z}_p$ are cosets of $p^n \mathbb{Z}_p$ (these form a basis for the $p$-adic topology on $\mathbb{Z}_p$), and $\nu$ maps each such coset to an open set in the inverse limit. Therefore $\nu$ is a topological ring isomorphism $\mathbb{Z}_p \cong \varprojlim \mathbb{Z}_p / p^n \mathbb{Z}_p$. [/step] [step:Pass to the inverse limit of $\mathbb{Z}/p^n\mathbb{Z}$ using the quotient isomorphisms] By the theorem [Ideals of $\mathbb{Z}_p$ and Quotients](/theorems/???), the canonical maps \begin{align*} \phi_n: \mathbb{Z}/p^n\mathbb{Z} \xrightarrow{\;\sim\;} \mathbb{Z}_p / p^n \mathbb{Z}_p \end{align*} are ring isomorphisms for each $n \geq 1$. These isomorphisms are compatible with the transition maps: the diagram \begin{align*} \mathbb{Z}/p^{n+1}\mathbb{Z} &\xrightarrow{\phi_{n+1}} \mathbb{Z}_p / p^{n+1}\mathbb{Z}_p \\ \downarrow & \qquad\qquad \downarrow \\ \mathbb{Z}/p^n\mathbb{Z} &\xrightarrow{\phi_n} \mathbb{Z}_p / p^n\mathbb{Z}_p \end{align*} commutes (both paths send $a + p^{n+1}\mathbb{Z}$ to $a + p^n\mathbb{Z}_p$). By the functoriality of inverse limits, the $\phi_n$ induce an isomorphism of topological rings \begin{align*} \varprojlim_{n} \mathbb{Z}/p^n\mathbb{Z} \cong \varprojlim_{n} \mathbb{Z}_p / p^n \mathbb{Z}_p. \end{align*} Composing with $\nu$, we obtain the topological ring isomorphism \begin{align*} \mathbb{Z}_p \cong \varprojlim_{n} \mathbb{Z}/p^n\mathbb{Z}. \end{align*} [/step]