Definition of exponential function, single-valued or multi-valued? If we define
$$e^z=1+z+\frac{z^2}{2!}+\cdots$$
then it is single-valued.
However, if we write
$$e^z=e^{z\ln e}$$
then it is multi-valued.
Besides, $a^z$ is multi-valued in general. It is kind of strange if only when the base is $e$ that it is single-valued.
My thought:
Is it true that there are two exponential functions, let's call them $\exp(z)$ and $e^z$?
Where $\exp(z)$ is defined by
$$\exp(z)=1+z+\frac{z^2}{2!}+\cdots$$
and is single-valued, while $e^z$ is defined by
$$e^z = \text{exp}(z\ln e)$$
and is multi-valued?
Here $\ln z$ is defined by $\exp(\ln z)=z$ and is multi-valued.

It's true that the complex map $\ln$ is multi-valued, but in the case of your second definition it doesn't matter, because $\exp$'s periodicity kills the extra arguments of $\ln$, so the definition becomes single-valued and perfectly ok.
For example, $\arg(e)=\Theta=0$ and $\ln|e|=1$, so your definition becomes:
$$e^{z\cdot \ln(e)}=e^{z\cdot (\ln|e|+(\Theta+2k\pi))i}\text{  },k\in\mathbb{Z}\Rightarrow$$
$$e^{z\cdot \ln(e)}=e^{z(1+2k\pi i)}\Rightarrow$$
$$e^{z\cdot \ln(e)}=e^z\cdot e^{z2k\pi i},k\in\mathbb{Z}\text{ (1)}$$
Correction after egreg's comment:
The following two steps are in fact wrong.
$$e^{z\cdot \ln(e)}=e^z\cdot (e^{2k\pi i})^z\Rightarrow$$
$$e^{z\cdot \ln(e)}=e^z\cdot 1^z=\exp(z)$$
They would be right, had the OP specified only the principal branch of $\ln$. Since there's no such concensus, I am correcting my answer, based on egreg's comment.
I agree then that the definition is multi-valued, and it seems the actual values are given by the last line (1).
