Here is how we explain the resolution in Bayesian Essentials with R rephrased for your problem (but you can find the same approach in most textbooks):
With each $x_i$ is associated a missing variable $z_i$ [your $\theta_i$] that indicates its component.
Formally, this means that we have a hierarchical structure associated
with the model:
$$
z_i|\pi\sim\mathscr{B}(\pi)
$$
and
$$
x_i|z_i=1,\beta \sim \beta x^{\beta-1}\qquad x_i|z_i=0,\beta \sim \mathcal{U}(01,1)
$$
The completed likelihood corresponding to the missing structure is such that
$$
\ell(\beta,\pi|\mathbf{x},\mathbf{z})=\prod_{i=1}^n \pi^{z_i}(1-\pi)^{1-z_i}\,(1-x_i)^{(\beta-1) z_i}\beta^{z_i}
$$
and the corresponding posterior is
$$
\pi(\beta,\pi|\mathbf{x},\mathbf{z}) \propto \beta^{-.999}\exp(-0.001\beta)\,\prod_{i=1}^n \pi^{z_i}(1-\pi)^{1-z_i}\,(1-x_i)^{(\beta-1) z_i}\beta^{z_i},
$$
where $\mathbf{z}=(z_1,\ldots,z_n)$.
As I dislike very much the BUGS type priors with almost zero
parameters, I will now use $\beta^{-1}$ instead of
$\beta^{-.999}\exp(-0.001\beta)$
Using this completion, the joint posterior distribution of $(\beta,\pi)$ can be written in closed form as
$$
f(\beta,\pi|\mathbf{x},\mathbf{z}) \propto \pi^{\sum_{i=1}^nz_i}(1-\pi)^{n-\sum_{i=1}^nz_i}\,\beta^{\sum_{i=1}^n z_i-1}\left[\prod\nolimits_{z_i=1}(1-x_i)\right]^{\beta-1},
$$
If $\pi$ and $\beta$ are independent a priori, then, given $\mathbf{z}$,
the vectors $\pi$ and $\mathbf{x}$ are independent; that is,
$f(\pi|\mathbf{x},\mathbf{z})=f(\pi|\mathbf{z})$. Moreover, in that case, $\beta$ is also independent a posteriori from $\pi$ given $\mathbf{x}$ and $\mathbf{z}$, with density $f(\beta|\mathbf{x},\mathbf{z})$. If we apply the Gibbs sampler in this problem, it involves the successive simulation of $\mathbf{z}$ and $(\pi,\beta)$ conditional on one another and on the data:
Initialization:
Choose $\pi^{(0)}$ and $\beta^{(0)}$ arbitrarily.
Iteration $t$ $(t\ge 1)$:
- For $i=1,\ldots,n$, generate $z_i^{(t)}$ such that $$\mathbb{P}\left(z_i=1\right)\propto \pi^{(t-1)}f\left(x_i|\beta^{(t-1)},z_i\right)$$
- Generate $\pi^{(t)}$ according to $f(\pi|\mathbf{z}^{(t)})$.
- Generate $\beta^{(t)}$ according to $f(\beta|\mathbf{z}^{(t)},\mathbf{x})$.
Try to solve the full conditionals before reading further
In your specific case, you can derive the full conditional $f(\pi|\mathbf{z})$ and $f(\beta|\mathbf{z},\mathbf{x})$ from the joint above:
$$\eqalign{
f(\beta,\pi|\mathbf{x},\mathbf{z}) &\propto \pi^{\sum_{i=1}^nz_i}(1-\pi)^{n-\sum_{i=1}^nz_i}\,\beta^{\sum_{i=1}^n z_i-1}\left[\prod\nolimits_{z_i=1}(1-x_i)\right]^{\beta-1}\\
&\propto f(\pi|\mathbf{z}) f(\beta|\mathbf{x},\mathbf{z})\\
}$$
If you separate the terms in $\pi$ and the terms in $\beta$ you get $$f(\pi|\mathbf{z})\propto\pi^{\sum_{i=1}^n z_i}(1-\pi)^{n-\sum_{i=1}^nz_i}$$ which corresponds to a Beta $$\mathcal{B}e\left(1+\sum\nolimits_{i=1}^n z_i,1+n-\sum\nolimits_{i=1}^n z_i\right)$$ distribution and $$f(\beta|\mathbf{x},\mathbf{z})\propto\beta^{\sum_{i=1}^n z_i-1}\left[\prod\nolimits_{z_i=1}(1-x_i)\right]^{\beta}= \beta^{\sum_{i=1}^n z_i-1} \exp\left[-\beta\sum\nolimits_{z_i=1} \log(1-x_i)\right] $$ which corresponds to a Gamma $$\mathcal{Ga}\left(\sum\nolimits_{i=1}^n z_i,\sum\nolimits_{z_i=1} \log(1-x_i)\right)$$ distribution.
