I need help solving the following textbook problem (translated):
Suppose $z_1,z_2,...,z_n$ (where $n\geq2$) are complex numbers. Show that: $$\int_\gamma \frac{1}{(z-z_1)(z-z_2)...(z-z_n)}dz=0$$ for every simple closed curve $\gamma$ that encloses all $z_1,...,z_n$.
My initial idea is to use Cauchy's integral formula with partial fraction decomposition. The integrand equals $$\frac{A}{z-z_1}+\frac{B}{z-z_2}+\frac{C}{z-z_3}...$$ and with Cauchy's integral formula the integral evaluates to $$2\pi(A+B+C...).$$
I find it hard to show that $A+B+C...$ would be $0$. Help is appreciated!