How did they get this proof for CDF of gamma distribution? Let $$T \sim Gamma(\alpha, \lambda)$$
$$f(t) = \frac1{ \Gamma(\alpha)}{\lambda^\alpha}t^{\alpha-1}{e^{-\lambda t}} \qquad t,\alpha,\lambda > 0$$
The CDF result :
$$F(t) = 1 - \sum_{i=0}^{\alpha-1}{\frac {(\lambda t)^i} {i!}e^{-\lambda t}}, \qquad t,\alpha,\lambda > 0$$
or
$$F(t) =e^{-\lambda t}\sum_{i=0}^{\alpha-1}{\frac {(\lambda t)^i} {i!}}, \qquad t,\alpha,\lambda > 0$$
 A: This discrete summation works only for integer-valued $\alpha$, and there's a reason to that. That reason is itself the perspective of what the identity expresses (it's an expression for real), rather than merely an algebraic manipulation.
$F(t) = P(T_{\alpha} < t)$ is the probability that $\alpha$ occurrence of the underlying Poisson process takes place before time $t$. 
Here I attached the subscript $\alpha$ for $T$ as an index to emphasize the duality that will be clear soon.
In other words, up to time $t$, there are at least $\alpha$ occurrences. Therefore, consider the random variable $X_t$ that follows the discrete Poisson distribution (counts of occurrences). 
$$P( X_t = i) = e^{-\lambda t} \frac{ (\lambda t)^i }{ i !}$$
It is subscripted (indexed) by $t$ to indicate that it is over time interval $[0, t]$, with $E[X_t] = \lambda t$. Namely, $(\lambda t)$ as a whole is the "parameter" of the Poisson distribution.
$$F_T(t) \equiv P(T_{\alpha} < t) = P( X_t \geq \alpha) \\
\implies F_T(t)  = 1 - P( X_t < \alpha)$$
which is the desired expression.
A: It is the series expansion of the CDF. For $T \sim \text{Gamma}(a,λ)$, the standard CDF is the regularized Gamma $Γ$ function :
$$F(x;a,λ) = \int_0^x f(u;a,λ)\mathrm{d}u= \int_0^x \frac1{ \Gamma(a)}{\lambda^a}t^{a-1}{e^{-\lambda u}}\mathrm{d}u =  \frac{γ(a,λx)}{Γ(α)}$$
where $γ$ is the lower incomplete gamma function.
If α is a positive integer (i.e., the distribution is an Erlang distribution), the cumulative distribution function has the following series expansion: 
$$F(x;a,λ) = 1 - \sum_{i=0}^{a-1}{\frac {(\lambda x)^i} {i!}e^{-\lambda x}} = e^{-\lambda t}\sum_{i=0}^{a-1}{\frac {(\lambda x)^i} {i!}}$$
