Let $z\in\mathbb{C}$ and $r\in\mathbb{R}$. Assuming that $\exp ^r z$ can be a multi-valued function (and $\exp rz$ cannot), there always exists (for any given $z$ and $r$) some value of $\exp ^r z$ such that $$\exp ^r z=\exp rz.$$ How I could I prove this?
As a starting point, I'm using the fact that $\exp z=\sum_{n\ge 0}\frac{z^n}{n!}$ for all $z\in\mathbb{C}$.
I'm motivated by the following proof that, for all $w,z\in\mathbb{C}$, $\exp w\exp z=\exp (w+z)$: $$\begin{align}\exp w\exp z&=\sum_{n\ge 0}\frac{w^n}{n!}\sum_{k\ge 0}\frac{z^k}{k!}=\sum_{n\ge 0}\sum_{0\le k\le n}\frac{w^k}{k!}\frac{z^{n-k}}{(n-k)!}\\&=\sum_{n\ge 0}\frac{1}{n!}\sum_{0\le k\le n}\frac{n!}{k!\, (n-k)!}w^k z^{n-k}=\sum_{n\ge 0}\frac{(w+z)^n}{n!}\\&=\exp (w+z).\end{align}$$
Maybe the proof of the relation in my question could be done in a similar fashion.
Since $\exp w\exp z=\exp(w+z)$, the cases $r\in\Bbb N$ follow by induction, and $r=0$ is trivial, and $\exp-w\exp w=1$ extends to $r\in\Bbb Z$. We get to $\in\Bbb Q$ by writing $r=p/q$ with $p\in\Bbb Z,\,q\in\Bbb N$ so$$(\exp^rz)^q=\exp^pz=\exp pz=\exp(qrz)=(\exp rz)^q.$$We extend to $r\in\Bbb R$ by continuity in $r$: for any sequence of rationals $r_n$ with $\lim_{n\to\infty}r_n=r$,$$\exp^rz=\lim_{n\to\infty}\exp^{r_n}z=\lim_{n\to\infty}\exp(r_nz)=\exp rz.$$