Given $\Theta = \theta$, let $X_1, X_2, \dots, X_n, X_{n+1} \sim \mathcal{N}(\theta, \sigma^2)$ be independent.
$\Theta \sim \mathcal{N}(\theta_0, \tau^2)$.
What is the easiest way to find the distribution of $X_{n+1}\mid X_1, \dots, X_n$?
I am looking only for hints. How I would start this is $$f_{X_{n+1} \mid X_1, \dots, X_n}(t \mid x_1, \dots, x_n) = \dfrac{f_{X_1, \dots, X_{n+1}}(x_1, \dots, x_{n}, t)}{f_{X_1, \dots, X_{n}}(x_1, \dots, x_n)}$$ and conditioned on $\Theta$, we know that the $X$s are independent, so $$f_{X_1, \dots, X_{n+1}}(x_1, \dots, x_{n}, t) = \int_{-\infty}^{\infty}f_{X_1, \dots, X_{n+1}\mid \Theta}(x_1, \dots, x_n, t\mid \theta)\pi(\theta)\text{ d}\theta$$ and we can write $$f_{X_1, \dots, X_{n+1}\mid \Theta}(x_1, \dots, x_n, t\mid \theta) = f_{X_1 \mid \Theta}(x_1\mid\theta)\cdots f_{X_n\mid \Theta}(x_n\mid \theta)f_{X_{n+1}\mid \Theta}(t\mid \theta)$$ by independence, and similarly for the denominator, but this looks disgusting.
Hints:
If you need to write out the conditional density of $X_{n+1}$ given the rest, it might help to use