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.
 A: The notation $e^z$ is just shorthand for $\exp(z)$. Everybody agrees that when $e^z$ is evaluated it means
$$
e^z=\exp(z)=\sum_{n\ge0}\frac{z^n}{n!}
$$
Notation is not always consistent, unfortunately.
Indeed, if we would interpret $e^z$ as $\exp(z\log e)$, where $\log e$ is any determination of the logarithm, we'd get infinitely many values, unless $z$ is an integer. This is because $\log e$ can be any complex number $w$ such that $\exp(w)=e$ and it's easy to see that $w=1+2ki\pi$, for $k\in\mathbb{Z}$. Thus $\exp(zw)=\exp(z(1+2ki\pi))$. For instance, if $z=1/2$, we would have to assign $e^{1/2}$ both the values $\sqrt{e}$ and $-\sqrt{e}$.
However, as I said at the beginning, notation is a bit sloppy in this respect. Writing $e^z$ instead of $\exp(z)$ is deemed more practical and so $e^z$ used with the convention that it means the same as $\exp(z)$, mainly because single valued functions can be better manipulated algebraically.
As you note, people generally avoids using $a^z$ for $a\ne e$, when $z$ possibly varies in the complex numbers. In the case of positive real $a$, however, since $\log a$ has a well defined unique real value, there's no difficulty in defining and using $a^z=\exp(z\log a)$, where $\log a$ means that unique real value. The algebraic property $a^{z_1+z_2}=a^{z_1}a^{z_2}$ holds without restriction, with this convention (but only for positive real $a$).
A: *

*Indeed, unfortunately, the standard definitions of $e^z\;(z\in\mathbb
   C)$ as single-valued—which we generally want—are
consistent, when using the logarithmic $(\log)$ definition of $e^z$,
only with using the principal branch $\mathrm{Log}.$
I.e.,  usually,  $$e^z := 1+z+\frac{z^2}{2!}+\cdots \\=
   e^{z\,\mathrm{Log}(e)} \\≠  e^{z\log(e)};$$ however when wanting the
function to be multi-valued (e.g. when determining $n$th roots),
$$e^z := e^{z\log(e)}.$$


*The multi-valued definition of $e^z$ (i.e., using $\log$ instead
$\mathrm{Log}$ to define $e^z$) results in $$e^{x+iy} = 
   e^{x+iy}e^{2kπ(ix-y)}$$  being multivalued—which is generally
not desired. For example, this apparently contradictory result
ensues: $$e^{i\pi}=-e^{2n\pi^2}\\\neq-1.$$


*Until reading this post, I had thought that $\exp()$ and $e^{()}$ are
synonymous; it turns out that the convention is that the former
denotes the single-valued variant of the latter. It's indeed a good
disambiguation.
A: You run into problems with the second definition of the exponential, as the function $e^z$ on the complex plane is not 1-1.  In $\mathbb{R}$, it is true that $e^x=e^y$ implies $x=y$, but for example, $e^0=1=e^{2\pi i}$ in $\mathbb{C}$ and $0 \neq 2\pi i$.  So $e^z$ does not have an inverse--hence your second equation being multi-valued.
A: Define
\begin{eqnarray}
\ln z &=& \text{Ln}|z| + i \arg z\\
      &=& \text{Ln}|z| + i(\arg z + 2 \pi n)
\end{eqnarray}
Where $\text{Ln}$ is the ordinary real logarithm, clearly the complex logarithm is multi-valued as it depends on $n$.
A: 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).
