It is easy to show that the generating function of the consecution $\text{Cyc}(\mathcal{Z})$ is $\ln\left(\frac{1}{1-z}\right)$.
My question is, how to start from here to prove that the generating function for $\text{Cyc}_{\geq r}(\mathcal{Z})$ is
$$\sum_{n\geq r}(n-1)!\frac{z^n}{n!}=\ln\left(\frac{1}{1-z}\right)-\sum_{n=1}^{r-1}\frac{z^n}{n}$$
I suppose that to pass to that expression I must do some algebraic step but I can't see it. ($\mathcal{Z}$ is a atomic class)
Simply note that the exponential generating function for ${\rm Cyc}_n(\mathcal{Z})$ is given by $$C_n(z)=(n-1)!\frac{z^n}{n!}=\frac{z^n}n$$ so the exponential generating function for ${\rm Cyc}_{\geq r}(\mathcal{Z})$ is given by $$\sum_{n\geq r}C_n(z)=\sum_{n\geq r}\frac{z^n}{n}=\sum_{n\geq 1}\frac{z^n}{n}-\sum_{n=1}^{r-1}\frac{z^n}{n}=\log\left(\frac{1}{1-z}\right) -\sum_{n=1}^{r-1}\frac{z^n}{n}$$ as desired.