Exercise: Suppose $X_1,\ldots,X_n|\Theta = \theta \stackrel{iid}{\sim} \operatorname{Pois}(\theta)$ and $\Omega\sim\operatorname{Ga}(\alpha, \beta)$. Derive the posterior distribution and compute the posterior mean.
What I've tried: I know that the posterior distribution is equal to $$f_{\Theta|X}(\theta|x) = \dfrac{f_{X|\Theta}(x|\theta)\,f_\Theta(\theta)}{\int f_{X|\Theta}(x|\theta)\,f_\Theta(\theta)d\theta}$$ so with $f_{X|\Theta}(x|\theta)\,f_{\Theta}(\theta) = \dfrac{\theta^{\sum x_i}e^{-n\theta}}{\prod x_i !}\dfrac{\beta^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\beta\theta}$ this means that $$f_{\Theta|X}(\theta|x)=\dfrac{\dfrac{\theta^{\sum x_i}e^{-n\theta}}{\prod x_i !}\dfrac{\beta^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\beta\theta}}{\displaystyle\int\dfrac{\theta^{\sum x_i}e^{-n\theta}}{\prod x_i !}\dfrac{\beta^\alpha}{\Gamma(\alpha)}\theta^{\alpha-1}e^{-\beta\theta}d\theta}.$$ I think this is some sort of Beta distribution. However, I'm not sure which one and how to get there.
Question: How do I solve this exercise?
Thanks in advance!
You've mostly got this; the key to the problem is to ignore irrelevant information. Right now, you're trying to keep track of too much information you don't need -- which makes the problem much harder than it needs to be.
The denominator of the RHS of your first equation is some constant that doesn't depend on $\theta$ -- it doesn't matter what constant it is, since ultimately you know it will be whatever constant is required to turn the numerator into a proper pdf. So instead of working with: $$ f_{\Theta|X}(\theta|x) = \frac{f_{X|\Theta}(x|\theta) f_\Theta(\theta) } {\int f_{X|\Theta}(x|\theta)f_\Theta(\theta)d\theta}$$ you should be working with $$ f_{\Theta|X}(\theta|x) \propto f_{X|\Theta}(x|\theta) f_\Theta(\theta) $$ that is, all you need to know is that the LHS is proportional to the RHS, i.e. they are equal up to some normalizing constant. As you have found: $$f_{\Theta|X}(\theta|x) \propto \theta^{\sum x_i}e^{-n\theta}\theta^{\alpha-1}e^{-\beta\theta}$$ Notice that I left off not only the denominator from the equation you found in your question, I left off a bunch of stuff from the numerator, too. Why? Because all the stuff I left out is constant with respect to $\theta$, and all we care about is the distribution of $\theta$. So all that other stuff is again just part of the normalizing constant for the pdf we care about.
Simplifying the above, we get: $$f_{\Theta|X}(\theta|x) \propto \theta^{\alpha+\sum x_i-1}e^{-(n+\beta)\theta}$$
Now, what kind of distribution has $f(x) = c \cdot x^{a} e^{-bx}$ for some constant $c$? (Hint: it's not Beta, as you suggested.) Once you know that, you can easily figure out what $c$ should be in this case -- but really, who cares what the normalizing constant is, since you will already know the exact distribution (including parameters) of $\Theta|X$. And once you have the distribution, it will be trivial to recover the posterior mean.