To find the probability function for the gamma distribution, once again focus on the development of F(x). Assuming r is a natural number greater than 1 and noting that X measures the interval length needed in order to achieve the rth success
\begin{align*}
F(x) & = P(X \le x)\\
& = 1 - P(X \gt x)\\
& = 1 - P(\text{fewer than r successes in [0,x]})\\
& = 1 - \big [ \frac{(\lambda x)^0 e^{-\lambda x}}{0!} + \frac{(\lambda x)^1 e^{-\lambda x}}{1!} + ... + \frac{(\lambda x)^{r-1} e^{-\lambda x}}{(r-1)!} \big ]\\
& = 1 - \sum_{k=0}^{r-1} \frac{(\lambda x)^k e^{-\lambda x}}{k!}
\end{align*}
where the discrete Poisson probability function is used on the interval [0,x]. The derivative of this function however is "telescoping" and terms cancel. Indeed,
\begin{align*}
F'(x) & = \lambda e^{-\lambda x}/0!\\
& - \lambda e^{-\lambda x}/1! + \lambda x \cdot \lambda e^{-\lambda x}/1!\\
& - \lambda^2 2x e^{-\lambda x}/2! + \lambda^2 x^2 \cdot \lambda e^{-\lambda x}/2!\\
& - \lambda^3 3x^2 e^{-\lambda x}/3! + \lambda^3 x^3 \cdot \lambda e^{-\lambda x}/3!\\
& . . .\\
& - \lambda^{r-1} (r-1)x^{r-2} e^{-\lambda x}/(r-1)! + \lambda^{r-1} x^{r-1} \cdot \lambda e^{-\lambda x}/(r-1)!\\
& = \lambda^r x^{r-1} e^{-\lambda x}/(r-1)!
\end{align*}
where you can replace \((r-1)! = \Gamma(r)\text{.}\)