The typical approach when one sees a function which seems to have a pole of degree $n$ at $z_0$ is to check $$\lim_{z\rightarrow z_0}(z-z_0)^n f(z)$$ If the limit exists and is non-zero, we have a pole of degree $n$.
I have never seen a proof/formal statement of this, so I wanted to ask whether it is just obvious (enough), and whether we need to check some condition before using this reasoning to check for a pole.
If $n$ is a positive integer and $\lim_{z \to z_0} (z-z_0)^{n} f(z)$ exists and is not zero then $f$ has a pole of order $n$ at $z_0$. No further condition is necessary. Proof of this is based on the fact that $(z-z_0)^{n} f(z)$ has a removable singularity at $z_0$.