On slide 23 of these slides, it is stated that an $n$ body phase space element $d\Phi_n(P; p_1, \ldots, p_n)$ may be decomposed according to the recurrence relation
\begin{align*} \mathrm{d} \Phi_{n}\left(P ; p_{1}, \ldots, p_{n}\right) &=\mathrm{d} m_{12 \ldots(n-1)}^{2} \mathrm{~d} \Phi_{2}\left(P ; p_{12 \ldots(n-1)}, p_{n}\right) \\ &\times \mathrm{d} \Phi_{n-1}\left(P ; p_{1}, \ldots, p_{(n-1)}\right) \tag{1} \end{align*}
where $P$ is the 4-momentum of the initial state with total mass $P^2 = M^2$ and
$$ m^2_{12\ldots n} \stackrel{\text{def}}{=} (p_1 + p_2 +\cdots +p_n)^2. $$
On the following slide, the result of $n=4$ is given by
\begin{align*} \mathrm{d} \Phi_{4}\left(P ; p_{1}, p_{2}, p_{3}, p_{4}\right) &\propto \frac{\sqrt{\lambda\left(M^{2} ; m_{4}^{2}, m_{123}^{2}\right)}}{M^{2}} m_{123} \mathrm{~d} m_{123} \mathrm{~d} \Omega_{1234} \\ &\times \frac{\sqrt{\lambda\left(m_{123}^{2} ; m_{3}^{2}, m_{12}^{2}\right)}}{m_{123}^{2}} m_{12} \mathrm{~d} m_{12} \mathrm{~d} \Omega_{123} \\ &\times \frac{\sqrt{\lambda\left(m_{12}^{2} ; m_{1}^{2}, m_{2}^{2}\right)}}{m_{12}^{2}} \mathrm{~d} \Omega_{12}\tag{2} \end{align*}
where $\lambda$ is the Källén fucnction and we have used the fact that
$$ \mathrm{d} \Phi_{2}\left(P ; p_{1}, p_{2}\right) \propto \frac{\sqrt{\lambda\left(M^{2}, m_{1}^{2}, m_{2}^{2}\right)}}{M^{2}} \mathrm{~d} \Omega_{12} $$
Question: I cannot see how equation (2) is derived from equation (1). It appears that the $n=4$ case is able to be reduced into a product of $d\Phi_2(P; p_{123}, p_{4})$, $d\Phi_2(p_{123}; p_{12}, p_{3})$ and $d\Phi_2(p_{12}; p_{1}, p_{2})$, but I don't see how this can be if $d\Phi_{n-1}$ always has $P$ as the first argument in (1). Is this simply a typo or am I missing something?