I am proving that Poisson distribution is a difference of two Gamma CDFs. I used integration by parts and the property of gamma function $\Gamma(k+1) = k! = k \Gamma(k)$, but could not get to the form I want.
Let $T_k \sim \text{Gamma}(k)$ and $N_t \sim \text{Poisson}(t)$. Suppose we know the following equality $$ \begin{align} P(N_t) = P(T_k \leq t < T_{k+1}) = P(T_k \leq t) - P(T_{t+1} \leq t). \end{align} $$
I want to show (from the textbook): $$ \begin{align} P(N_t=k) = \frac{1}{\Gamma{(k)}} \int_0^t e^{-x} x^{k-1} dx - \frac{1}{\Gamma(k+1)} \int_0^t e^{-x} x^{k} dx = e^{-t} \frac{t^k}{k!}. \end{align} $$
For the first term, I have: $$ \begin{align} &\quad\frac{1}{\Gamma{(k)}} \int_0^t e^{-x} x^{k-1} dx \\ &= \frac{1}{\Gamma{(k)}} \left\{ -e^{-t} \cdot t^{k-1} + (k-1) \int_0^t e^{-x} x^{k-2} dx \right\} \quad \text{Integration by parts}\\ &= \frac{1}{k \Gamma{(k)}} \left\{ k ( -e^{-t} \cdot t^{k-1}) +k(k-1) \int_0^t e^{-x} x^{k-2} dx \right\} \\ &= \frac{1}{k \Gamma{(k)}} \left\{ k ( -e^{-t} \cdot t^{k-1}) +k! \right\}. \\ \end{align} $$
Similarly, the second term will become $$ \begin{align} &\quad\frac{1}{\Gamma{(k+1)}} \int_0^t e^{-x} x^{k} dx \\ &= \frac{1}{k\Gamma{(k)}} \left\{ -e^{-t} \cdot t^{k} + k! \right\}. \\ \end{align} $$
However, taking the difference of these two does not become the form I want.
Keep the first integral as it is:
$$ \begin{align} \frac{1}{k \Gamma(k)} \left[ k\int_0^t e^{-x} x^{k-1} dx \right] - \frac{1}{\Gamma(k+1)} \left[ -e^{-t} t^k + k \int_0^t e^{-x} x^{k-1} dx \right] \end{align} $$
This gives you a Poisson distribution.