I'd like to prove that $\frac{e^{z_1}}{e^{z_2}}=e^{z_1-z_2}$. Obviously this is true for real numbers, but here, $z_1=x_1+iy_1$ and $z_2=x_2+iy_2$, so it needs to be proven.
$$\frac{e^{z_1}}{e^{z_2}}=\frac{e^{x_1}(\cos(y_1)+i\sin(y_1))}{e^{x_2}(\cos(y_2)+i\sin(y_2))}=\frac{e^{x_1}}{e^{x_2}}*\frac{(\cos(y_1)+i\sin(y_1))}{(\cos(y_2)+i\sin(y_2))}$$
I can ignore $\frac{e^{x_1}}{e^{x_2}}$ for now and try to prove that $$\frac{(\cos(y_1)+i\sin(y_1))}{(\cos(y_2)+i\sin(y_2))}=cos(y_1-y_2)+i\sin(y_1-y_2)$$
Then, because $\frac{e^{x_1}}{e^{x_2}}=e^{x_1-x_2}$ ($x_1$ and $x_2$ are real), I would know that $e^{x_1-x_2}=\cos(y_1-y_2)+i\sin(y_1-y_2)$ which is $e^{z_1-z_2}$.
I think this middle step that I don't know how to do can be done with the trig identities, $\cos(\alpha - \beta)=\cos \alpha \cos \beta + \sin \alpha \sin \beta$ and $\sin(\alpha - \beta)=\sin \alpha \cos \beta -\cos \alpha \sin \beta$. I just don't see how to make that work.
I think there is a simpler approach.
If you know that
$e^{x+iy}=e^xe^{iy}$
then in general you know that
$e^{z_1+z_2}=e^{z_1}e^{z_2}$
so
$e^{z_1-z_2}e^{z_2}=e^{z_1-z_2+z_2}=e^{z_1}$
and so
$e^{z_1-z_2}=\frac{e^{z_1}}{e^{z_2}}$