[proofplan] We treat the $p$-adic logarithm and exponential separately. For $\log(1+x) = \sum_{n=1}^\infty (-1)^{n+1} x^n/n$, we show the general term tends to zero by bounding the $p$-adic valuation of $x^n/n$ from below using $v(n) \leq \log_p n$, which grows sublinearly while $n \cdot v(x)$ grows linearly. For $\exp(x) = \sum_{n=0}^\infty x^n/n!$, we use Legendre's formula $v_p(n!) = (n - s_p(n))/(p-1)$ to show $v(x^n/n!) \to \infty$ precisely when $v(x) > 1/(p-1)$. Continuity of both maps follows from uniform convergence on compact sub-discs. [/proofplan]

[step:Establish convergence of $\log(1+x)$ for $|x| < 1$]Define the $p$-adic logarithm by the formal power series \begin{align*} \log(1+x) := \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}. \end{align*} Let $v = -\log_p |\cdot|$ denote the valuation on $K$ extending $v_p$, so that $|x| < 1$ is equivalent to $v(x) > 0$. We must show the general term $(-1)^{n+1} x^n / n$ tends to zero in $|\cdot|$, i.e., that $v(x^n/n) \to \infty$. We have $v(x^n/n) = n \cdot v(x) - v(n)$. Since $n$ is a positive integer and $v = v_p$ on $\mathbb{Q}$, we use the bound $v_p(n) \leq \log_p n$ (because $p^{v_p(n)}$ divides $n$, so $p^{v_p(n)} \leq n$). Therefore \begin{align*} v\!\left(\frac{x^n}{n}\right) = n \cdot v(x) - v_p(n) \geq n \cdot v(x) - \log_p n. \end{align*} Since $v(x) > 0$, the linear term $n \cdot v(x)$ dominates $\log_p n$ as $n \to \infty$, so $v(x^n/n) \to \infty$. By the [Convergence via Term Decay](/theorems/???) criterion for non-archimedean series, the series converges.[/step]

[guided]The $p$-adic logarithm is defined by the same formal power series as the real logarithm: \begin{align*} \log(1+x) := \sum_{n=1}^\infty (-1)^{n+1} \frac{x^n}{n}. \end{align*} In the archimedean setting, this converges for $|x| < 1$ by the alternating series test. In the non-archimedean setting, we have a simpler convergence criterion: a series $\sum a_n$ converges if and only if $|a_n| \to 0$. So we need to show $|x^n/n|$ tends to zero. Let $v = -\log_p |\cdot|$ be the valuation extending $v_p$. The condition $|x| < 1$ translates to $v(x) > 0$. For the $n$-th term: \begin{align*} v\!\left(\frac{x^n}{n}\right) = n \cdot v(x) - v_p(n). \end{align*} How large can $v_p(n)$ be? Since $p^{v_p(n)}$ divides $n$, we have $p^{v_p(n)} \leq n$, giving $v_p(n) \leq \log_p n$. This is a crude but sufficient bound. The lower estimate becomes \begin{align*} v\!\left(\frac{x^n}{n}\right) \geq n \cdot v(x) - \log_p n. \end{align*} Since $v(x) > 0$ is a positive constant and $\log_p n$ grows sublinearly, $n \cdot v(x) - \log_p n \to \infty$ as $n \to \infty$. The terms of the series tend to zero, so the series converges by the non-archimedean convergence criterion. Note that this argument does not require a sharp bound on $v_p(n)$. The sublinear growth of $v_p(n)$ compared to the linear growth of $n \cdot v(x)$ is all that matters.[/guided]

[step:Establish convergence of $\exp(x)$ for $|x| < p^{-1/(p-1)}$ using Legendre's formula]Define the $p$-adic exponential by \begin{align*} \exp(x) := \sum_{n=0}^\infty \frac{x^n}{n!}. \end{align*} We need $v(x^n/n!) \to \infty$. The key tool is Legendre's formula for the $p$-adic valuation of factorials: \begin{align*} v_p(n!) = \frac{n - s_p(n)}{p - 1}, \end{align*} where $s_p(n) = \sum_{i \geq 0} a_i$ is the sum of the base-$p$ digits of $n = \sum_{i \geq 0} a_i p^i$. Since $1 \leq s_p(n) \leq n$ for $n \geq 1$ and $s_p(n) \leq (p-1)(\lfloor \log_p n \rfloor + 1)$, we have the bounds \begin{align*} \frac{n-1}{p-1} \geq v_p(n!) \geq \frac{n}{p-1} - \lfloor \log_p n \rfloor - 1. \end{align*} In particular, $v_p(n!)/n \to 1/(p-1)$ as $n \to \infty$. Now \begin{align*} v\!\left(\frac{x^n}{n!}\right) = n \cdot v(x) - v_p(n!) = n \cdot v(x) - \frac{n - s_p(n)}{p-1} = n\!\left(v(x) - \frac{1}{p-1}\right) + \frac{s_p(n)}{p-1}. \end{align*} Since $s_p(n) \geq 1$ for $n \geq 1$, the second term is non-negative. If $v(x) > 1/(p-1)$, equivalently $|x| < p^{-1/(p-1)}$, then $v(x) - 1/(p-1) > 0$, and \begin{align*} v\!\left(\frac{x^n}{n!}\right) \geq n\!\left(v(x) - \frac{1}{p-1}\right) \to \infty. \end{align*} The series converges by the non-archimedean convergence criterion. Moreover, $v(x^n/n!) \geq 0$ for all $n \geq 0$ (since $n \cdot v(x) \geq n/(p-1) > (n - s_p(n))/(p-1) = v_p(n!)$), so each partial sum lies in $\mathcal{O}_K$, and the limit $\exp(x)$ lies in $\mathcal{O}_K$.[/step]

[guided]The exponential series $\exp(x) = \sum_{n=0}^\infty x^n/n!$ involves factorials, and the $p$-adic valuation of $n!$ is the decisive quantity. Unlike the logarithm, where the denominator $n$ has slowly growing valuation, the factorial $n!$ has valuation growing linearly in $n$, creating a race between $v(x^n) = n \cdot v(x)$ and $v(n!)$. **Legendre's formula.** The $p$-adic valuation of $n!$ is given by \begin{align*} v_p(n!) = \sum_{k=1}^\infty \left\lfloor \frac{n}{p^k} \right\rfloor = \frac{n - s_p(n)}{p-1}, \end{align*} where $s_p(n)$ is the digit sum of $n$ in base $p$. This is a classical result proved by counting the multiples of $p$, $p^2$, $p^3$, etc., up to $n$. **The critical threshold.** Since $s_p(n) \geq 1$ for $n \geq 1$, we have $v_p(n!) \leq (n-1)/(p-1)$. More usefully, $v_p(n!)/n \to 1/(p-1)$. So the $n$-th term of the exponential has valuation \begin{align*} v\!\left(\frac{x^n}{n!}\right) = n \cdot v(x) - \frac{n - s_p(n)}{p-1} = n\!\left(v(x) - \frac{1}{p-1}\right) + \frac{s_p(n)}{p-1}. \end{align*} For this to tend to $+\infty$, we need the dominant term $n(v(x) - 1/(p-1))$ to be eventually positive, which requires $v(x) > 1/(p-1)$, i.e., $|x| < p^{-1/(p-1)}$. **Image in $\mathcal{O}_K$.** When $v(x) > 1/(p-1)$, the entire Witt expansion of each term satisfies $v(x^n/n!) \geq 0$: indeed, $n \cdot v(x) \geq n/(p-1) > (n - s_p(n))/(p-1) = v_p(n!)$ since $s_p(n) \geq 1$. So every partial sum lies in $\mathcal{O}_K$, and the limit $\exp(x)$ lies in $\mathcal{O}_K$. **Why the radius is $p^{-1/(p-1)}$ and not $1$.** Unlike the logarithm, the exponential's denominator $n!$ has valuation growing as $n/(p-1)$, not sublinearly. The "cost" of dividing by $n!$ is roughly $n/(p-1)$ in valuation, which must be exceeded by the "gain" $n \cdot v(x)$. This forces $v(x) > 1/(p-1)$, a strictly smaller disc than $v(x) > 0$.[/guided]

[step:Verify continuity of both maps via uniform convergence on compact sub-discs] For the logarithm: let $0 < r < 1$ and consider the closed disc $D = \{x \in K : |x| \leq r\}$. On $D$, the $n$-th term satisfies $|x^n/n| \leq r^n / |n|_p^{-1}$. Since $r < 1$, the bound $r^n$ decreases geometrically while $|n|_p^{-1}$ grows at most polynomially, so the series converges uniformly on $D$. As $\mathbb{Z}_p$ is compact and a uniform limit of continuous functions is continuous, $\log(1+x)$ is continuous on $\{x : |x| < 1\}$. For the exponential: let $0 < r < p^{-1/(p-1)}$ and consider $D = \{x \in K : |x| \leq r\}$. On $D$, the $n$-th term satisfies \begin{align*} \left|\frac{x^n}{n!}\right| \leq r^n \cdot p^{(n - s_p(n))/(p-1)} = \left(r \cdot p^{1/(p-1)}\right)^n \cdot p^{-s_p(n)/(p-1)}. \end{align*} Since $r \cdot p^{1/(p-1)} < 1$, the dominant factor decreases geometrically, so the series converges uniformly on $D$. The exponential map is therefore continuous on $\{x : |x| < p^{-1/(p-1)}\}$. [/step]