Let \(z_0\in\CC\), let \(r>0\), and let \(f\) be holomorphic on the punctured disc \(D_r(z_0)\setminus\{z_0\}\). For an integer \(n\ge1\), \(f\) has a pole of order \(n\) at \(z_0\) if there exist \(0<\rho\le r\) and a holomorphic function \(h\colon D_\rho(z_0)\to\CC\) with \(h(z_0)\neq0\) such that \[ f(z)=(z-z_0)^{-n}h(z)\qquad\text{for } 0<\abs{z-z_0}<\rho. \] \(f\) has a pole at \(z_0\) if it has a pole of some order \(n\ge1\) there, and a pole of order \(1\) is a simple pole.
Let \(f\) be holomorphic on \(D_r(z_0)\setminus\{z_0\}\) and let \(n\ge1\). The following are equivalent.
\(f\) has a pole of order \(n\) at \(z_0\).
There is \(0<\rho\le r\) such that \(f\) has no zeros in \(D_\rho(z_0)\setminus\{z_0\}\) and \(g\coloneqq1/f\), extended by \(g(z_0)\coloneqq0\), is holomorphic on \(D_\rho(z_0)\) with a zero of order $n$ at \(z_0\).
The Laurent expansion of \(f\) on \(D_r(z_0)\setminus\{z_0\}\) has the form \(f(z)=\sum_{k\ge-n}c_k(z-z_0)^k\) with \(c_{-n}\neq0\).
Moreover, \(f\) has a pole at \(z_0\) if and only if \(\abs{f(z)}\to\infty\) as \(z\to z_0\).
(a)\(\Rightarrow\)(b). Since \(h(z_0)\neq0\) and \(h\) is continuous, shrinking \(\rho\) makes \(h\) nonvanishing on \(D_\rho(z_0)\). Then \(1/f(z)=(z-z_0)^n\cdot(1/h(z))\) for \(0<\abs{z-z_0}<\rho\), and the right side is holomorphic on \(D_\rho(z_0)\), vanishes at \(z_0\), and has \(1/h(z_0)\neq0\).
(b)\(\Rightarrow\)(a). Write \(g(z)=(z-z_0)^n k(z)\) with \(k\) holomorphic near \(z_0\) and \(k(z_0)\neq0\); shrinking \(\rho\), \(k\) is nonvanishing, and \(h\coloneqq1/k\) satisfies \(f(z)=(z-z_0)^{-n}h(z)\) with \(h(z_0)\neq0\).
(a)\(\Rightarrow\)(c). Expand \(h(z)=\sum_{j\ge0}b_j(z-z_0)^j\) on \(D_\rho(z_0)\) with \(b_0=h(z_0)\neq0\). Then \(f(z)=\sum_{k\ge-n}b_{k+n}(z-z_0)^k\) on \(D_\rho(z_0)\setminus\{z_0\}\), and by uniqueness of Laurent coefficients this is the Laurent expansion of \(f\), with \(c_{-n}=b_0\neq0\).
(c)\(\Rightarrow\)(a). The power series \(h(z)\coloneqq\sum_{k\ge-n}c_k(z-z_0)^{k+n}\) converges for \(0<\abs{z-z_0}<r\), hence on \(D_r(z_0)\), and defines a holomorphic function there with \(h(z_0)=c_{-n}\neq0\) and \(f(z)=(z-z_0)^{-n}h(z)\).
Finally, if \(f\) has a pole of order \(n\), then \(\abs{f(z)}=\abs{h(z)}/\abs{z-z_0}^n\to\infty\) because \(h(z_0)\neq0\). Conversely, if \(\abs{f(z)}\to\infty\), then \(f\) has no zeros on some \(D_\rho(z_0)\setminus\{z_0\}\), and \(1/f\) is holomorphic and bounded there with \(1/f(z)\to0\). By Riemann's removable singularity theorem, \(g\coloneqq1/f\) with \(g(z_0)\coloneqq0\) is holomorphic on \(D_\rho(z_0)\). It is not identically zero, so \(z_0\) is a zero of \(g\) of some order \(n\ge1\), and (b) holds.