From the above, I follow the argument and see that the posterior distribution of $\theta$ is proportional to a pdf which has density of a beta random variable. But i do not understand why because the posterior is proportional to a distribution, implies it is exactly that distribution?
In Bayesian inference one often finds the posterior distribution as proportional to the product of the prior distribution and likelihood function. In case the prior and likelihood are 'conjugate' (mathematically compatible), it is possible to ignore constants.
Specifically, if the prior is Beta and the Likelihood is Binomial, the posterior will be Beta. As in your example, all that is necessary is to find the appropriate power of $\theta$ and of $(1-\theta).$
Once you know that the 'kernel' of the posterior (PDF without constant) is of the form $\theta^{1-\alpha_n}(1-\theta)^{1 - \beta_n},$ then you know that the posterior distribution is $\mathsf{Beta}(\alpha_n,\, \beta_n)$. If for some reason, you need to know the 'constant of integration' $K$ you can find it as $1/K = \int_0^1 \theta^{1-\alpha_n}(1-\theta)^{1 - \beta_n}\,d\theta,$ so that $\int_0^1 K \theta^{1-\alpha_n}(1-\theta)^{1 - \beta_n}\,d\theta = 1.$ Of course, $K = \frac{\Gamma(\alpha + \beta)}{\Gamma(\alpha)\Gamma(\beta)}.$ In short, there really can be only one PDF that has kernel $\theta^{1-\alpha_n}(1-\theta)^{1 - \beta_n},$ for $\alpha_n, \beta_n > 0.$
The conjugate distribution pair Poisson-gamma is used similarly for Bayesian inference on the Poisson rate parameter $\lambda.$