Beta-Binomial conjugacy

The likelihood of observing $k$ successes in $n$ trials given $\theta$ is $L(k\mid\theta)=\binom{n}{k}\theta^k(1-\theta)^{n-k}$. By the posterior-proportionality theorem above, $\pi(\theta\mid k)\propto L(k\mid\theta)\pi(\theta)=\binom{n}{k}\theta^k(1-\theta)^{n-k}\cdot\dfrac{\theta^{\alpha-1}(1-\theta)^{\beta-1}}{B(\alpha,\beta)}$.

The factors $\binom{n}{k}$ and $B(\alpha,\beta)$ do not depend on $\theta$, so they can be absorbed into the proportionality: $\pi(\theta\mid k)\propto\theta^{k}(1-\theta)^{n-k}\cdot\theta^{\alpha-1}(1-\theta)^{\beta-1}=\theta^{(\alpha+k)-1}(1-\theta)^{(\beta+n-k)-1}$.

This last expression is exactly the kernel (the $\theta$-dependent part) of a $\mathrm{Beta}(\alpha+k,\beta+n-k)$ density. Since a probability density on $(0,1)$ with this kernel has a unique normalizing constant (namely $1/B(\alpha+k,\beta+n-k)$, by definition of the Beta function), the posterior must be exactly $\pi(\theta\mid k)=\dfrac{\theta^{(\alpha+k)-1}(1-\theta)^{(\beta+n-k)-1}}{B(\alpha+k,\beta+n-k)}$, i.e. $\theta\mid k\sim\mathrm{Beta}(\alpha+k,\beta+n-k)$.