TITLE: Zero of the exponential p-adic
QUESTION [6 upvotes]: Consider the $p$-adic exponential defined over $\mathbb C_p$. One knows $\exp$ is analytic in the domain $\mathcal D=\{z\in\mathbb C_P\mid v_p(z)>\frac1{p-1}\}$. Does it exist an element $z_0\in\mathcal D$ such that $\exp(z_0)=0$?
Thanks in advance.

REPLY [8 votes]: It’s hard to see in what sense it could be true that the exponential function is “defined over $\Bbb C_p$”, since the logarithm is defined on the whole open unit disk there, and has so very many zeros.
If you look closely, you can see that for all $z\in\mathcal D$, we have $v_p(e^z - 1)=v_p(z)$. This obtains quite independently of any multiplicativity of the exponential.