Let $w$ be an integer. Then
$$
P(X\geqslant w)=1-P(X\leqslant w-1)=1-e^{-\lambda}\sum_{k=0}^{w-1}\frac{\lambda^k}{k!}.
$$
Now, we use that if $\Gamma(a,b)$ denotes the incomplete Gamma function, i.e.
$$
\Gamma(a,b)=\int_b^\infty t^{a-1}e^{-t}\,\mathrm dt,
$$
then
$$
\Gamma(n,b)=(n-1)!e^{-b}\sum_{k=0}^{n-1}\frac{b^k}{k!}
$$
provided that $n$ is an integer. Thus (recall that $\Gamma(n)=(n-1)!$ for $n$ integer)
$$
P(X\geqslant w)=1-\frac{\Gamma(w,\lambda)}{\Gamma(w)}.
$$
Rewriting this expression we arrive at the desired expression
$$
\begin{align}
P(X\geqslant w)&=\frac{\Gamma(w)-\Gamma(w,\lambda)}{\Gamma(w)}=\frac{1}{\Gamma(w)}\left(\int_0^\infty t^{w-1}e^{-t}\,\mathrm dt-\int_\lambda^\infty t^{w-1}e^{-t}\,\mathrm dt\right)\\
&=\frac{1}{\Gamma(w)}\int_0^\lambda t^{w-1}e^{-t}\,\mathrm dt=P(Y\leqslant\lambda).
\end{align}
$$
Just continue step (3).
\begin{align}
f_{Y \mid X}(y,i) &= \frac{f_{X,Y}(i,y)}{f_X(i)} \\
&= \large\frac{\frac{Ce^{-(\alpha + 1)y} \cdot
y^{s+i-1}}{i!}}
{\int_0^\infty \frac{Ce^{-(\alpha + 1)y} \cdot
y^{s+i-1}}{i!} \,dy} \\
&= \underbrace{\frac{1}{\int_0^\infty e^{-(\alpha + 1)y} \cdot
y^{s+i-1} \,dy}}_{\text{constant independent of }y} \cdot
e^{-(\alpha + 1)y} \cdot y^{s+i-1}
\end{align}
This is a constant multiplied by $e^{-(\alpha + 1)y} \cdot y^{s+i-1}$.
Recall that gamma distribution with parameters $(s+i,\alpha+1)$ has density $C_{(s+i,\alpha+1)} e^{-(\alpha + 1)y} \cdot y^{s+i-1}$ for all $y>0$. By the very definition of density function,
$$\int_0^\infty C_{(s+i,\alpha+1)} e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy = 1.$$
Observe that
$$\int_0^\infty \frac{1}{\int_0^\infty e^{-(\alpha + 1)y} \cdot
y^{s+i-1} \,dy} \cdot
e^{-(\alpha + 1)x} \cdot x^{s+i-1} \,dx
= \frac{\int_0^\infty e^{-(\alpha + 1)x} \cdot
x^{s+i-1} \,dx}{\int_0^\infty e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy}=1.$$
So $C_{(s+i,\alpha+1)} = \dfrac{1}{\int_0^\infty e^{-(\alpha + 1)y} \cdot y^{s+i-1} \,dy}$ and $f_{Y \mid X}$ is the density function of $\Gamma(s+i,\alpha+1)$.
Best Answer
Use the tower rule: $$ \mathbb{P}\left(X=n\right) = \mathbb{E}_\lambda\left(\underbrace{\mathbb{P}\left(X=n | \lambda\right)}_{X|\lambda \sim \operatorname{Poi}\left(\lambda\right)}\right) $$ Write out the probability for the Poisson random variable, and then write the definition of the expectation.
The final expectation can also be computed easily: $$ \mathbb{P}\left(X=n\right) = \int_0^\infty \frac{\lambda^n}{n!} \mathrm{e}^{-\lambda} \frac{\beta^\alpha}{\Gamma(\alpha)} \lambda^{\alpha-1} \mathrm{e}^{-\beta \lambda} \mathrm{d} \lambda = \binom{n+\alpha-1}{n} \left(\frac{\beta}{\beta+1} \right)^\alpha \frac{1}{(\beta+1)^n} $$ This shows that $X$ is a negative binomial random variable, $X \sim \operatorname{NB}\left(\alpha, \frac{\beta}{\beta+1}\right)$.