Let $p$ be a prime number and denote by $R(f)$ the radius of convergence of a power series $f(x) \in \mathbb{C}\_p[[x]]$, where $\mathbb{C}\_p$ is the completion of the algebraic closure of $\mathbb{Q}\_p$, the field of $p$-adic numbers. Given two power series $f(x), g(x) \in \mathbb{C}\_p[[x]]$, it is known that the radius of convergence of the product $h(x) = f(x)g(x)$ is at least the minimum of the radius of convergence of the two series $f(x)$ and $g(x)$. In other words, we have
$$ R(h) \geq \min\{R(f), R(g)\}.$$
Keep in mind $(1-x)(1+x+x^2+\dots) = 1$ as an example for the strict inequality. Is there a way to easily predict when $R(h) > \min\{R(f), R(g)\}$ and find $R(h)$ explicitly? More specifically, I'm interested in computing the radius of convergence of power series of the form $\exp(f(x))$ for $f(x) \in x\mathbb{C}\_p[[x]]$.


For example, let $f(x) = \exp(x)$ and $g(x) = \exp(x^p/p)$. Then $R(f) = R(g) = (1/p)^{1/(p-1)}$ and using the fact that the Artin-Hasse exponential series $$\text{AH}(x) = \exp(x + x^p/p + x^{p^2}/p^2 + \cdots)$$ 
lies in $\mathbb{Z}\_p[[x]]$ (which implies $R(\text{AH}) \geq 1$), $h(x) = \exp(x + x^p/p)$ has radius of convergence 
$$ R(h) = R\left(\exp\left(\frac{x^{p^2}}{p^2}\right)\right) = \left(\frac{1}{p}\right)^{\frac{(2p-1)}{p^2(p-1)}}> \left(\frac{1}{p}\right)^{\frac{1}{p-1}}. $$
The importance of this example comes from the fact that if we set $\pi$ to be a root of $x+x^p/p = 0$, then $h(\pi)$ is a non-trivial $p$-th root of unity in $\mathbb{C}\_p$. This provides an analytic representation of $p$-th roots of unity, exploited in particular in Dwork's proof of the rationality of zeta functions over finite fields. 


More generally, using this method one can show that for any $n \geq 1$, we have
$$ R(\exp(x+x^p/p+\cdots + x^{p^n}/p^n)) = R(\exp(x^{p^{n+1}}/p^{n+1})). $$
Even though I understand the details involved in this calculation, I don't know if there is some more general theory underlying these examples. It may be helpful to share similar examples that you know. For instance, is it always the case that 
$$ R(\exp(f(x))) > R(\exp(x)),$$
 given $f(x) \in x\mathbb{C}\_p[[x]]$ has a nonzero root $\alpha \in \mathbb{C}\_p$ of absolute value $R(\exp(x)) = (1/p)^{1/(p-1)}$ and no non-zero roots of smaller absolute value? 



When does the radius of convergence of the product of two $p$-adic power series increase?
Here is a counterexample to your question at the end, for each $p$. Let $f\_u(x) = x + ux^p/p$ for $u \in \mathbf C\_p$ with $|u|\_p = 1$ and $|u-1|\_p = 1$. (Such $u$ can be taken in $\mathbf Z\_p^\times$ if $p > 2$, but you need to go outside $\mathbf Q\_p$ if $p = 2$ to an extension with residue field of size greater than $2$.) Since $|u|\_p = 1$, all the nonzero roots of $f\_u(x)$ in $\mathbf C\_p$ have absolute value $(1/p)^{1/(p-1)} = R(\exp)$.


To find the $p$-adic radius of convergence of $\exp(f\_u(x))$, write
$$
\exp(f\_u(x)) = \exp\left(x+\frac{x^p}{p}\right)\exp\left((u-1)\frac{x^p}{p}\right)
$$
as formal power series. On the right side, the first factor has radius of convergence greater than $(1/p)^{1/(p-1)}$, as you noted. 


Since $|u-1|\_p = 1$, the second factor has radius of convergence equal to that of $\exp(x^p/p)$, which is $(1/p)^{1/(p-1)}$. The reciprocal $(\exp(x + x^p/p))^{-1}$ has the same radius of convergence as $\exp(x+x^p/p)$, even for $p=2$, so $\exp(f\_u(x))$ has radius of convergence equal to $(1/p)^{1/(p-1)}$. This is the counterexample to your question. 


