Let $E$ be a normed $\mathbb R$-vector space$^1$ and $\ln$ denote the principal branch of the complex logarithm.
We can show the following result:
Theorem: Let $\varphi:E\to\mathbb C\setminus\{0\}$ with $\varphi(0)=1$. If $\left.\varphi\right|_{\overline B_r(0)}$ is uniformly continuous and $$\inf_{x\in\overline B_r(0)}|\varphi(x)|>0\tag1$$ for all $r>0$, then there is a unique $f\in C(E,\mathbb C)$ with $f(0)=0$ and $e^f=\varphi$. Moreover, for every $n\in\mathbb N$, there is a unique $g_n\in C(E,\mathbb C\setminus\{0\})$ with $g_n(0)=1$ and $g_n^n=\varphi$; in fact, $$g_n=e^{\frac fn}\tag2.$$
By this result, the notation $$\varphi^{\frac1n}:=g_n\;\;\;\text{for }n\in\mathbb N$$ is well-defined.
Question: For $m,n\in\mathbb N$, it is tempting to write $\varphi^{\frac mn}$ instead of $g_n^m$. Are we able to justify this notation by showing that whenver $m,m',n,n'\in\mathbb N$ satisfy $\frac mn=\frac{m'}{n'}$, then $g_n^m=g_{n'}^{m'}$?
I wasn't able to show this result, but I'm really sure that it holds. My main problem is that it is intuitively so trivial that it's easy to make a stupid mistake.
I wasn't able to utilize this, but we may note $$\left(g_{mn}^m\right)=\varphi\tag3$$ and hence $$g_n=g_{mn}^m\tag4.$$
$^1$ If this generality is preventing you from providing an answer, feel free to assume $E=\mathbb R^d$ (in which case we may further assumpe that $\varphi$ is uniformly continuous (on the whole space) and hence $(1)$ is trivially satisfied by compactness of closed balls in $\mathbb R^d$).
You have that $g_n^m = (e^{\frac{f}{n}})^m = e^{\frac{m}{n}f}$, which obviously depends only of $\frac{m}{n}$. The main point here is that the same (unique) $f$ satisfies $g_n = e^{\frac{f}{n}}$ for every value of $n$.