I'm learning about Bayesian statistics and I've come across the following example:
Let \begin{split}X_1, ..., X_n \stackrel{i.i.d}{\sim} Ber\,(\theta),\\ \Theta \sim Be(\alpha, \beta). \end{split}
We have $$f_{\Theta|X}(\theta|x) = \dfrac{f_{X|\Theta}(x|\theta)\,f_{\Theta}(\theta)}{\int f_{X|\Theta}(x|\theta)\,f_{\Theta}(\theta)\,d\theta}$$
My book claims that in this case $\Theta |X \sim Be(\alpha + S,\beta + n -S)$. I tried to check this and I got the following result: $$f_{\Theta|X}(\theta|x) = \frac{\theta^{S+\alpha -1}(1-\theta)^{n- S + \beta -1}}{\int\theta^{S+\alpha -1}(1-\theta)^{n- S + \beta -1} d\theta},$$
where $S$ is the numbers of times that $X_i$ equals $1$, that is $S = \sum\limits_{i = 1}^n x_i$. I've my results are correct, I don't see how $\Theta |X \sim Be(\alpha + S,\beta + n -S)$. I understand that I'm missing something, since the exponentials of $\theta$ and $(1-\theta)$ in the fraction match the supposed distribution exactly.
Question: Why is $\Theta |X \sim Be(\alpha + S,\beta + n -S)$?