Zero of the exponential p-adic

The exponential function satisfies $\exp \left({x + y}\right) = \exp\left(x \right) \exp\left(y \right)$ for $x, y$ in the convergence domain. It also satisfies $\exp \left( 0 \right) = 1$. So if $\exp \left( z_0 \right) = 0$, then $0 = \exp \left( z_0 \right) \cdot \exp\left(-z_0 \right) = \exp \left( z_0 + (-z_0) \right) = \exp\left( 0 \right) = 1$, which is a contradiction.

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.