As every number theorist learns, the radius of convergence of $exp(x)$, defined by the usual power series in a neighborhood of zero, is $$\rho = p^{-1/(p-1)}.$$ This is typically proven by computing the $p$-adic absolute value of $n!$.
I imagine that this might also be proven by using the fact that the differential equation $Df - f = 0$ (to which $f(x) = exp(x)$ is a solution) has an irregular singular point at $\infty$. My intuition in the $p$-adic case, from looking at a paper of Bombieri and Dwork a long time ago, is that this irregular singular point "pushes" convergence away from infinity -- hence the relatively small radius of convergence of $exp(x)$ $p$-adically, near zero.
Is there a somewhat general statement along these lines -- that an irregular singular point of a differential equation at a point $s$ (on a smooth projective curve, let's say) will cause the (generic?) radius of convergence of a series solution to the differential equation at a point $t$ to be less than something (involving $s$, $t$, and invariants of irregular singular points, etc.)? Something that, when you "plug in" the differential equation $Df - f = 0$, and the points $s = \infty$ and $t = 0$, will output the radius of convergence of $exp(x)$?
It will indeed be interesting if this approach can predict the radius of convergence; but I like to look at the exponential as a secondary thing, the inverse of the $p$-adically much more fundamental logarithm. This is defined on the whole open unit disk, with derivative a unit, so $\exp$ can be defined only on an open disk of the same size as the one in which $\log$ is one-to-one, in other words on the disk $z\colon v(z)>1/(p-1)$. No need for memory here: sketch the Newton polygon of the log and see what (nonunit) homothetic conjugation needs to be applied to make the result integral. –  Lubin Jan 27 '11 at 21:44