There are several solutions to this problem but I am interested in the solution in Casella & Burger Pg. 100. The problem shows that if $X$ follows gamma($\alpha$, $\beta$), a random variable and $Y \sim 
 Poisson(x/\beta)$, then 
$P(X\le x) = P(Y\ge \alpha)$. 

In the text,
$P(X \le x) = \frac{1} {(\alpha - 2)! \beta^{\alpha - 1}}\int_{0}^{x} t^{\alpha - 2} e^{-t/\beta} \,dx - P(Y = \alpha -1 )$

After second integration by parts (as suggested by the text), I get 

$P(X \le x) = \frac{1} {(\alpha - 3)! \beta^{\alpha - 2}}\int_{0}^{x} t^{\alpha - 2} e^{-t/\beta} \,dx  - P(Y = \alpha -2 ) - P(Y = \alpha -1 )$

I surmise that at the $n^{th}$ step I will have 
$P(X \le x) = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx  -  P(Y = \alpha -(n-1)) \ldots -P(Y = \alpha -2 ) - P(Y = \alpha -1 )$

which will be 

$\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx  - [ P(Y = \alpha -(n-1)) \ldots +P(Y = \alpha -2 ) + P(Y = \alpha -1 )]$

$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx - [ P(Y = \alpha -(n-1)) \ldots +P(Y = \alpha -2 ) + P(Y = \alpha -1 )]$

$ = \frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx  -  P(Y \le \alpha- 1) $

I want to argue that as $n \rightarrow \infty $ , $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta}= 1 $. In which case,
$ = 1  -  P(Y \le \alpha- 1) = P(Y \ge \alpha) $. 

I am not sure however, if $\frac{1} {(\alpha - n)! \beta^{\alpha - (n-1)}}\int_{0}^{x} t^{\alpha - (n-1)} e^{-t/\beta} \,dx$  is an integral of a valid pdf. 




Thanks for the help in advance.