Now I do not understand and could not solve beyond this. The author says that the CDF can be simplified by applying 2.321.2 from the following attached image (reference: **Table of Integrals, Series, and Products**):

The result (which is a CDF from $0$ to $g$) that author has yield is :

$$ F_{f(g)}(g) = e^{-K}\sum_{k=0}^\infty \frac{K^{k}}{k!} \Bigg\{e^{-g}\sum_{m=0}^k(-1)^{2m+1}m! \binom{m}{k} x^{k-m} + k!\Bigg\}$$

I really cannot understand how he came up with this.

My result yields the following

$$F_{f(g)}(g) = e^{-K}\sum^\infty_{k=0} \sum^k_{m=0} \frac{K^k}{k!} {-e^{-g}g^{k-m}(e^{-g} - 1)}$$

Any idea?