Let $f$ be a complex function with $n$ number of $1,2,\dots,n$ order poles: $z_1,z_2,\dots,z_n$ such that: $f(z)=\frac{1}{(z-z_1)(z-z_2)^2\dots(z-z_n)^n}$and $\gamma$ a simple closed curve which contains all of the singularities. Using the Cauchy's integral formula:
$$\oint_\gamma f(z)dz=2\pi i\bigg(\frac{1}{1!} \big[(z-z_1)f(z)\big]\bigg|_{z=z1}+\frac{1}{2!}\ \big[(z-z_2)^2f(z)\big]^{\prime}\bigg|_{z=z_2}+\dots+\frac{1}{n!}\ \big[(z-z_n)^nf(z)\big]^{(n-1)}\bigg|_{z=z_n}\bigg)=2\pi i\sum_{k=1}^n\frac{1}{k!}\big[(z-z_k)^kf(z)\big]^{(k-1)}\bigg|_{z=z_k}$$
Is this true?