For every complex $z$, $e^z \ne 0$ In Real and Complex Analysis, 3rd Edition, Walter Rudin advances the following: $?

If there was a $z\in \mathbb{C}$ such that $e^z=0$ we would have

$e^z\cdot e^{-z}=0\cdot e^{-z} \Rightarrow\\\1=0$

which simply is not true.

_EDIT_

Another way to look at it is by using the form $e^z=e^{x+iy}=e^x(\cos y+i\sin y)$

with $x,y \in \mathbb{R}$

Thus $0=e^{x+iy}=e^x(\cos y+i\sin y)\Rightarrow \cos y=-i\sin y$

which cannot hold.