for any prime $p \in \mathbb{N}$ use the corresponding symbol $q$ to denote the quantity $$q=p^{\frac1{p-1}}$$ and for $n \in \mathbb{N}$ define: $$ Q_n= \prod_{p \le n} q $$ empty products evaluate to unity.
let $\sigma_p(n)$ denote the sum of the digits in the $p$-ary notation for $n$ and define $$n_{\sigma} =\prod_{p \le n} q^{\sigma_p(n)} $$ then: $$ e^z = \sum_{n=0}^{\infty} n_{\sigma}\left(\frac{z}{Q_n}\right)^n \tag{1}$$
follows because $$ n! = \prod_{p \le n}p^{\frac{n-\sigma_p(n)}{p-1}} $$ Q. the dissociation of the terms of the series using $n_{\sigma}$ and $Q_n$ seems superficially "natural" because of the way $Q_n$, like the variable $z$ is raised to the $n^{th}$ power in the corresponding term. however this procedure arteficially introduces a surd into the notation whose appearances alway cancel out. I wondered if there are other contexts in which $\sqrt[p-1]{p}$ appears ?
In fact $p^{-1/(p-1)}$ is the radius of convergence of the so-called $p$-adic exponential function on $\Bbb C_p{}^\dagger$, and additionally sorta-kinda shows up in the formula for the discriminant of a cyclotomic field: $\Delta(\Bbb Q(\zeta_n)/\Bbb Q)=(in\prod_{p\mid n}\color{Blue}{p^{-1/(p-1)}})^{\varphi(n)}$. (In fact, I've been meaning to ask someone if these two appearances are connected in some algebro-geometric way or are only coincidental.)
${}^\dagger$Secretly this is the same example as yours.