Given that
$$ \frac{d^nf(z)}{dz^n} \bigg|_{z=z_0}=\frac{n!}{2\pi i}\oint \frac{f(z)}{(z-z_0)^{n+1}}dz $$
Can we apply the same formula to a product of the form $g(z) = z^mf(z)$? For instance
$$ \frac{d^n }{dz^n}[z^mf(z)]\bigg|_{z=z_0}=\frac{n!}{2\pi i}\oint \frac{z^mf(z)}{(z-z_0)^{n+1}}dz $$
The one you've written is correct because $z^mf(z)$ is holomorphic if you know that $f$ is holomorphic on the given domain.