Well it's obvious if you consider it a definition. However if you've defined $\displaystyle e^x$ as the power series $\displaystyle e^x=1+x+\frac{x^2}{2!}+...$ then it's less obvious. It's possible to prove, using the power-series definition, that $\displaystyle e^xe^y = e^{x+y}$, using an essentially combinatorial argument (which I can supply if you wish). It also follows from the power-series definition that $\displaystyle e^0=1$. From these two identities it follows that
$\displaystyle e^ze^{-z}=e^{z-z}=e^0=1$
So that $\displaystyle e^{-z}$ is the multiplicative inverse of $\displaystyle e^z$, i.e. $\displaystyle e^{-z}=(e^z)^{-1}$. Note also that the above shows that $\displaystyle e^z$ is never 0, because it is always invertible.
Hope that helps!