How do you take the product of Bernoulli distribution? I have a prior distribution,
$$p(\boldsymbol\theta|\pi)=\prod\limits_{i=1}^K p(\theta_i|\pi).$$
$\theta_i$ can equal $0$ or $1$, so I am using a Bernoulli distribtion so that
$$p(\boldsymbol\theta|\pi)=\prod\limits_{i=1}^K \pi^{\theta_i}(1-\pi)^{1-\theta_i}.$$
I then want to add this distribution onto my marginal likelihood to make up my posterior. Should I solve it as
$$p(\boldsymbol\theta|\pi)=\pi^{K\theta_i}(1-\pi)^{K(1-\theta_i)} \, \, ?$$
But then is the product of bernoulli distributions the binomial distribution?
Then should my answer be
$$p(\boldsymbol\theta|\pi)=\left(\begin{array}c K\\ t \end{array}\right)\pi^{t}(1-\pi)^{K-t)} $$
where $K$ is the maximum number of $\theta_i$'s allowed, and $t=\{0, 1\}$  , (i.e. $t=0\, \, \text{or}\, \, 1$)?
What form do I add this prior to my likelihood?

The equation you have can be represented as follows:
$$p(\boldsymbol x|\theta)=\prod\limits_{i=1}^K \theta^{x_i}(1-\theta)^{1-x_i}=\theta^{\sum_i x_i}(1-\theta)^{K-\sum_i x_i}$$
We have the Bayes rule
$$p(\theta|x)=\frac{p(x|\theta)p(\theta)}{p(x)}$$
as $\theta$ is known, we have the joint density $p(x,\theta)=p(\theta,x)$ which specifies all the information we need.
