Residue formula at a pole of order $m$

Step 1 (Write the Laurent series explicitly). Since $f$ has a pole of order $m$ at $z_0$, its Laurent series has the form $f(z) = \dfrac{a_{-m}}{(z-z_0)^m} + \cdots + \dfrac{a_{-1}}{z-z_0} + a_0 + a_1(z-z_0) + \cdots$ with $a_{-m} \neq 0$.

Step 2 (Clear the pole). Multiply both sides by $(z-z_0)^m$: $g(z) := (z-z_0)^m f(z) = a_{-m} + a_{-m+1}(z-z_0) + \cdots + a_{-1}(z-z_0)^{m-1} + a_0(z-z_0)^m + \cdots$, which is an ordinary power series, so $g$ is holomorphic at $z_0$.

Step 3 (Recognize the coefficient as a derivative). In this power series, $a_{-1}$ is the coefficient of $(z-z_0)^{m-1}$. For any holomorphic $g(z) = \sum_j c_j (z-z_0)^j$, Taylor's formula gives $c_j = \dfrac{g^{(j)}(z_0)}{j!}$, so $a_{-1} = \dfrac{g^{(m-1)}(z_0)}{(m-1)!}$.

Step 4 (Conclude). Since $\mathrm{Res}(f,z_0) = a_{-1}$ and $g^{(m-1)}(z_0) = \lim_{z\to z_0} \dfrac{d^{m-1}}{dz^{m-1}}[(z-z_0)^m f(z)]$, substituting gives $\mathrm{Res}(f, z_0) = \dfrac{1}{(m-1)!} \lim_{z \to z_0} \dfrac{d^{m-1}}{dz^{m-1}} \left[ (z-z_0)^m f(z) \right]$. Setting $m=1$ collapses the derivative and factorial to give $\mathrm{Res}(f, z_0) = \lim_{z \to z_0} (z-z_0) f(z)$.