Let \(f\) be holomorphic on a punctured disc \(D_r(z_0)\sm\ts{z_0}\) and have a pole at \(z_0\), so that \(\abs{f(z)}\to\infty\) as \(z\to z_0\). Then there exist a least integer \(n\ge1\) and a function \(h\) holomorphic on a neighborhood of \(z_0\) such that \[ f(z)=(z-z_0)^{-n}h(z) \] near \(z_0\), \(z\neq z_0\). For this \(n\), the function \(h\) is unique and \(h(z_0)\neq0\); the integer \(n\) is the order of the pole.
Since \(\abs{f}\to\infty\), there is \(0<\rho\le r\) with \(f\neq0\) on \(D_\rho(z_0)\sm\ts{z_0}\). The function \(g\coloneqq1/f\) is holomorphic and bounded there, so by Riemann’s removable singularity theorem it extends holomorphically to \(D_\rho(z_0)\) with \(g(z_0)=0\). The extension is not identically zero, so its zero at \(z_0\) has a finite order \(n\ge1\): \(g(z)=(z-z_0)^nk(z)\) with \(k\) holomorphic and nonvanishing on a neighborhood of \(z_0\). Then \(h\coloneqq1/k\) is holomorphic near \(z_0\), \(h(z_0)\neq0\), and \(f(z)=(z-z_0)^{-n}h(z)\).
If also \(f(z)=(z-z_0)^{-m}\tilde h(z)\) with \(\tilde h\) holomorphic near \(z_0\) and \(m<n\), then \(\tilde h(z)=(z-z_0)^{m-n}h(z)\) for \(z\neq z_0\), and \(\abs{\tilde h(z)}\to\infty\) as \(z\to z_0\) because \(h(z_0)\neq0\); this contradicts continuity of \(\tilde h\). So \(n\) is least, and \(h=(z-z_0)^nf\) is determined off \(z_0\), hence everywhere by continuity.