 Solution

A pole of order $n$ means $f(z)=z^{-n}g(z)$ with $g$ holomorphic near zero and $g(0)\ne 0$. Equivalently, the principal part is finite and starts with $\boxed{a_{-n}{z}^{-n},a_{-n}\ne 0}$, while $a_k=0$ for $k<-n$. In particular $n$ is the exact order, not merely an upper bound.