Conditional density following Gamma distribution We know that if a random variable say x ~ Gamma(a, b), then its probability density function is  $ \propto x^{a-1} exp^{-bx}$.
In a Bayesian hierarchical model, for example
$Z_1, \cdots, Z_n |\theta \sim iid \; Gamma(r, \theta)$,  (r known)$ ;
$\theta \sim \; Gamma(a, b)$,
the full conditional distribution of $\theta$ is
$p(\theta | Z_{1:n}) \propto p(\theta, Z_{1:n}) 
\propto p(Z_{1:n}|\theta) \times p(\theta)$,
by $Z_{1:n} | \theta \sim iid \; Gamma(r, \theta)$, I think $p(Z_{1:n}|\theta)$ shall be $\Pi_i Z_i^{r-1} exp^{- \theta Z_i}$, but the reading material suggests
$p(Z_{1:n}|\theta) = \Pi_i \theta^{r} exp^{-\theta Z_i}$,
I'm not sure I fully understand why it writes $\theta^{r}$ instead of $Z_i^{r-1}$ in front of $exp^{-\theta Z_i}$.
Can anyone explain? Thanks.

The full joint density of $Z_{1:n}$ conditional on $\theta$ is
$$
p(z_{1:n}|\theta)= \prod_{i=1}^n \frac{\theta^r}{\Gamma(r)} z_i^{r-1} \exp(-\theta z_i) \,.
$$
If we consider this as a function of $z_{1:n}$, the term $\theta^{r}/\Gamma(r)$ is just a normalising constant and can be dropped.
As you point out, the conditional distribution of $\theta$ given $z_{1:n}$ (i.e. the posterior of $\theta$ given the observed sample) is
$$
p(\theta|z_{1:n}) \propto p(\theta) p(z_{1:n}|\theta)  = p(\theta) \prod_{i=1}^n \frac{\theta^r}{\Gamma(r)} z_i^{r-1} \exp(-\theta z_i)
$$
This is a function of $\theta$, so we need to keep the terms involving $\theta$ (such as $\theta^{r}$), but we can drop the other terms (such as $z_i^{r-1}$). We can thus write
$$
p(\theta|z_{1:n}) \propto p(\theta) \prod_{i=1}^n \theta^r \exp(-\theta z_i)
\propto \theta^{a-1} \exp(-b\theta)\,\, \theta^{nr} \exp(-\theta n \bar{z})
$$
which we identify as another Gamma density.
