Showing the asymptotic behavior of the expectation of a function of a Galton-Watson process using the Stirling formula

Question

So right now I am trying to understand the proof in this paper. Let $(Z_n)_{n\in\mathbb{N}_0}$ be a Galton-Watson process with infinite offspring mean (I think supercriticality works as well), $\xi>0$ and $b>1$. The first statement of the paper that I do not understand is that \begin{align*} \int_0^{\infty}\exp(-\xi t)(1-\exp(-b^nt))^{Z_n}dt=\xi^{-1}\frac{\Gamma(1+\xi b^{-n})\Gamma(Z_n+1)}{\Gamma(Z_n+1+b^{-n}\xi)} \end{align*} I tried to transform the integral on the left into the beta function but it does not work properly. In addition the arguments in the gamma functions on the right do not seem to fit. Shouldnt the denominator be $\Gamma(Z_n+2+b^{-n}\xi)$? The second statement is that \begin{align*} \frac{\Gamma(Z_n+1)}{\Gamma(Z_n+1+b^{-n}\xi)}=(Z_n+1)^{-b^{-n}\xi}+o(1), \end{align*} where $o(1)$ is a random variable, which is uniformly bounded in $n$. I do not see why this should hold. In another paper I was reading, the author states that this was also unclear to him but he was able to derive that \begin{align*} E\left\lbrack \frac{\Gamma(Z_n+1)}{\Gamma(Z_n+1+b^{-n}\xi)}\right\rbrack\sim E\lbrack(Z_n+1)^{-b^{-n}\xi}\rbrack\text{ as }n\to\infty \end{align*} (which is also my main goal) by using Stirling's formula. Unfortunately, he did not give the proof for this. Can anybody help me?