[proofplan] We prove that every $p$-adic integer has a unique base-$p$ expansion and extend this to all non-zero elements of $\mathbb{Q}_p$. For the $\mathbb{Z}_p$ case, we invoke the topological ring isomorphism $\mathbb{Z}_p \cong \varprojlim \mathbb{Z}/p^n\mathbb{Z}$ established in the theorem [$\mathbb{Z}_p$ Is the $p$-Adic Completion of $\mathbb{Z}$](/theorems/???): a coherent sequence in the inverse limit determines and is determined by a sequence of digits $a_i \in \{0, 1, \ldots, p-1\}$. For $\mathbb{Q}_p^\times$, we factor out the appropriate power of $p$ to reduce to the $\mathbb{Z}_p$ case. [/proofplan] [step:Establish the unique $p$-adic expansion for elements of $\mathbb{Z}_p$] By the theorem [$\mathbb{Z}_p$ Is the $p$-Adic Completion of $\mathbb{Z}$](/theorems/???), the canonical map $\nu: \mathbb{Z}_p \to \varprojlim \mathbb{Z}/p^n\mathbb{Z}$ is an isomorphism of topological rings. Under this isomorphism, each $a \in \mathbb{Z}_p$ corresponds to a unique coherent sequence $(c_n)_{n \geq 1}$ with $c_n \in \mathbb{Z}/p^n\mathbb{Z}$ and $c_{n+1} \equiv c_n \pmod{p^n}$. Represent each $c_n$ by the unique integer $x_n \in \{0, 1, \ldots, p^n - 1\}$. The coherence condition $x_{n+1} \equiv x_n \pmod{p^n}$ means $x_{n+1} - x_n = a_n p^n$ for a unique digit $a_n \in \{0, 1, \ldots, p-1\}$, with $x_1 = a_0$. Inductively, $x_n = \sum_{i=0}^{n-1} a_i p^i$ is the base-$p$ representation of $x_n$. The series $\sum_{i=0}^{\infty} a_i p^i$ converges in $\mathbb{Z}_p$: since $|a_i p^i|_p \leq p^{-i} \to 0$ and $\mathbb{Q}_p$ is complete, convergence follows from the [Convergence via Term Decay](/theorems/???) theorem. Its limit equals $a$ because $|a - x_n|_p \leq p^{-n} \to 0$. **Uniqueness:** Suppose $a = \sum_{i=0}^{\infty} a_i p^i = \sum_{i=0}^{\infty} b_i p^i$ with $a_i, b_i \in \{0, \ldots, p-1\}$. Then for each $n \geq 1$, the partial sums $\sum_{i=0}^{n-1} a_i p^i$ and $\sum_{i=0}^{n-1} b_i p^i$ represent the same element of $\mathbb{Z}/p^n\mathbb{Z}$ (since both are congruent to $a$ modulo $p^n \mathbb{Z}_p$). Since these partial sums lie in $\{0, \ldots, p^n - 1\}$, they are equal as integers: $\sum_{i=0}^{n-1} a_i p^i = \sum_{i=0}^{n-1} b_i p^i$. Comparing the coefficient of $p^{n-1}$ (after subtracting the common lower terms) gives $a_{n-1} = b_{n-1}$. Since $n$ was arbitrary, $a_i = b_i$ for all $i \geq 0$. [guided] The existence of the $p$-adic expansion is really a restatement of the completeness isomorphism $\nu: \mathbb{Z}_p \xrightarrow{\sim} \varprojlim \mathbb{Z}/p^n\mathbb{Z}$, unpacked in terms of digits. Each element of the inverse limit is a compatible system of residues. Concretely, an element of $\varprojlim \mathbb{Z}/p^n\mathbb{Z}$ is a sequence of integers $x_1, x_2, x_3, \ldots$ with $x_n \in \{0, \ldots, p^n - 1\}$ and $x_{n+1} \equiv x_n \pmod{p^n}$. This compatibility condition is exactly the condition that the base-$p$ digits agree: $x_{n+1}$ is obtained from $x_n$ by appending one more digit at the top. Writing $x_n = a_0 + a_1 p + \cdots + a_{n-1} p^{n-1}$ with $a_i \in \{0, \ldots, p-1\}$, the coherent sequence is completely determined by the digit sequence $(a_0, a_1, a_2, \ldots)$. The corresponding element of $\mathbb{Z}_p$ is $a = \sum_{i=0}^{\infty} a_i p^i$. For uniqueness, the key point is that the digits are uniquely determined by the residues modulo $p^n$, and the residues are uniquely determined by $a$ (via the isomorphism $\nu$). Alternatively, if two expansions gave the same element, subtracting yields $\sum_{i=0}^{\infty} (a_i - b_i) p^i = 0$. Let $j$ be the smallest index with $a_j \neq b_j$. Then $|a_j - b_j|_p = 1$ (since $0 < |a_j - b_j| < p$) and $\left|\sum_{i=j}^{\infty} (a_i - b_i)p^i\right|_p = |p^j(a_j - b_j + p(\cdots))|_p = p^{-j}$ by the [Isoceles Triangle Principle](/theorems/???), contradicting the sum being zero. [/guided] [/step] [step:Extend to $\mathbb{Q}_p^\times$ by factoring out $p^n$] Let $a \in \mathbb{Q}_p^\times$. Set $n := v_p(a) = -\log_p |a|_p \in \mathbb{Z}$, so $|a|_p = p^{-n}$. Define $b := p^{-n} a$. Then \begin{align*} |b|_p = |p^{-n}|_p \cdot |a|_p = |p|_p^{-n} \cdot p^{-n} = p^{n} \cdot p^{-n} = 1, \end{align*} so $b \in \mathbb{Z}_p^\times \subset \mathbb{Z}_p$. By the first part of the proof, $b$ has a unique expansion \begin{align*} b = \sum_{i=0}^{\infty} a_i p^i, \quad a_i \in \{0, 1, \ldots, p-1\}. \end{align*} Since $|b|_p = 1$, we must have $a_0 \neq 0$ (otherwise $|b|_p \leq |p|_p = p^{-1} < 1$, contradicting $|b|_p = 1$). Multiplying by $p^n$: \begin{align*} a = p^n b = \sum_{i=0}^{\infty} a_i p^{i+n} = \sum_{i=n}^{\infty} a_{i-n} p^i. \end{align*} Relabelling $a_{i-n}$ as the digit in position $i$, we obtain \begin{align*} a = \sum_{i=n}^{\infty} a_i p^i, \quad a_i \in \{0, 1, \ldots, p-1\}, \quad a_n \neq 0. \end{align*} **Uniqueness** follows from the uniqueness of the expansion of $b = p^{-n} a \in \mathbb{Z}_p$: if $a = \sum_{i=n}^{\infty} a_i p^i = \sum_{i=n}^{\infty} b_i p^i$ with $a_n, b_n \neq 0$, then $p^{-n} a = \sum_{i=0}^{\infty} a_{i+n} p^i = \sum_{i=0}^{\infty} b_{i+n} p^i$, so $a_{i+n} = b_{i+n}$ for all $i \geq 0$. [/step]