When manipulating probabilities, it is very convenient to know some standard equalities, such as Bayes' theorem. However, it is also very important (perhaps even more so) to work out the dependencies between the involved variables.
The paper is actually simply using Bayes' theorem, but then simplifies the result considering said dependencies (or lack thereof).
A raw application of Bayes' would yield
$$
p(\lambda_1, \lambda_2, n|x_{1:N})=\frac{p(x_{1:N}|\lambda_1, \lambda_2, n)p(\lambda_1, \lambda_2, n)}{p(x_{1:N})} \propto p(x_{1:N}|\lambda_1, \lambda_2, n)p(\lambda_1, \lambda_2, n)
$$
First, let us examine the dependencies between the parameters. $\lambda_1, \lambda_2$ and $n$. These are drawn independently from a uniform and a Gamma distribution. Therefore,
$$
p(\lambda_1, \lambda_2, n) = p(\lambda_1) p(\lambda_2) p(n)
$$
Next, let us examine the term $p(x_{1:N}|\lambda_1, \lambda_2, n)$.
First, the $x_i$'s are mutually independent (wether or not we know $\lambda_1, \lambda_2, n$), as they are drawn independently. Therefore
$$
p(x_{1:N}|\lambda_1,\lambda_2,n)=\prod_{i=1}^N p(x_i|\lambda_1,\lambda_2,n)
$$
The next step is a bit more subtle. The value of $x_i$ depends on either $\lambda_1$ or $\lambda_2$, but not the two of them. Which one of the two influences $x_i$ is determined by $n$, specifically by whether or not $i\leq n$.
So we have
$$
p(x_i|\lambda_1,\lambda_2,n)=\begin{cases}
p(x_i|\lambda_1) & 1 \leq i \leq n \\
p(x_i|\lambda_2) & n < i \leq N
\end{cases}
$$
That is, the dependency on $n$ is "encoded" by $i$ and the depedency on either $\lambda_1$ or $\lambda_2$. In other words, $n$ does not influence the value $x_i$ any more than by determining which one of $\lambda_1,\lambda_2$ stays there, so once we choose one of the two, $n$ is no longer relevant.
Hence,
$$
p(x_{1:N}|\lambda_1,\lambda_2,n) = \prod_{i=1}^n p(x_i|\lambda_1)\prod_{i=n+1}^N p(x_i|\lambda_2)
$$
Finally, the products are collapsed into a joint distribution, presumably for notational simplicity (note that the independencies are employed right afterwards in the paper by expressing the joint in product form).
$$
\prod_{i=1}^n p(x_i|\lambda_1)=p(x_{1:n}|\lambda_1)
$$
$$
\prod_{i=n+1}^N p(x_i|\lambda_2)=p(x_{n+1:N}|\lambda_2)
$$
Putting it all together:
$$
p(\lambda_1, \lambda_2, n|X_{1:N}) \propto p(x_{1:n}|\lambda_1) p(x_{n+1:N}|\lambda_2) p(\lambda_1) p(\lambda_2) p(n)
$$
