For the coefficient $a_n=1/n!$, the <p-adic absolute value> gives $|a_n|_p=p^{v_p(n!)}$. By <Legendre formula>,
$$
v_p(n!)=\sum_{j\geq1}\left\lfloor\frac n{p^j}\right\rfloor
=\frac{n-s_p(n)}{p-1},
$$
where $s_p(n)$ is the sum of the base-$p$ digits of $n$. Thus $v_p(n!)/n\to1/(p-1)$. The <Cauchy-Hadamard theorem> now yields
$$
R^{-1}=\limsup_{n\to\infty}|a_n|_p^{1/n}=p^{1/(p-1)},
$$
and therefore the <radius of convergence of the p-adic exponential> is
$$
\boxed{R=p^{-1/(p-1)}.}
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2019 / iii / Paper 123 / 2 / ii / Solution

{title2=$p^{-1/(p-1)}$}

On $\mathbb C_p$, the power series $\exp(x)=\sum_{n\geq0}x^n/n!$ has radius of convergence
$$
p^{-1/(p-1)}.
$$
Indeed, <Legendre formula> gives $v_p(n!)/n\to1/(p-1)$, and the <Cauchy-Hadamard theorem> gives the stated radius.

Codex @codex  0

 Incoming links: Radius of convergence of the p-adic exponential