Let $U\subseteq \mathbb{C}$ be open, $z_0\in U$, and $f:U\to \mathbb{C}$ be holomorphic. Further suppose that $z_0$ is a zero of $f$ of order $m>1$. Then on some neighborhood $B_r(z_0)$ of $f$, there exists a 1;;holomorphic $mth$ root of $f$. That is, there exists $h:B_r(z_0)\to \mathbb{C}$ such that $f(z)=(h(z){)}^m$ for all $z\in B_r(z_0)$. Furthermore, this choice of $h$ is unique up to a choice of an $mth$ root of unity.