Let $\Omega \subseteq \mathbb C$ be a domain in the complex plane and $f \in \text {Hol} (\Omega).$ Let $z_0 \in \Omega$ be such that $f(z_0) \neq 0.$ Then for any $p \gt 0$ there exists a neighborhood $D$ (independent of the choice of $p \gt 0$) of $z_0$ such that there exists a holomorphic branch of $[f(z)]^p$ on $D.$
Since $f(z_0) \neq 0$ by neighborhood property there exists a ball $B(z_0, \rho) \subseteq \Omega$ around $z_0$ such that $f$ is non-vanishing on $B(z_0, \rho).$ Since open balls are simply connected, there exists a holomorphic function $g$ on $B(z_0, \rho)$ such that $\exp g(z) = f(z)$ for all $z \in B(z_0,\rho).$ Then by the holomorphic branch of $[f(z)]^p$ on $D$ does it mean the existence of the holomorphic function $z \mapsto \exp (p g(z))$ on $B(z_0,\rho),$ where $D = B(z_0,\rho)\ $?
It means that $f(z)^p$ is actually holomorphic. Note that $f(z)^p$ is defined as $e^{p\ln f(z)}$ where we need to choose a branch for the natural logarithm, $\ln$. We can only do this in a neighbourhood of $z_0$ if $f(z_0)\neq0$.