I'm working with the following definition of the octonions: $\mathbb{O} = \mathbb{H} \times \mathbb{H}$, endowed with the product $$(p,q)(r,s) = (pr - sq^*, p^*s + rq).$$ Conjugation is given by $(p,q)^* = (p^*, -q)$, and for each $P \in \mathbb{O}$, we define a norm $|P|^2 = P^*P$. I'm having a surprising amount of trouble showing that $|PQ| = |P||Q|$ for all $P,Q \in \mathbb{O}$. I'm aware of the following two identities: $(PQ^*)Q = P(Q^*Q)$ and $P^*(PQ) = (P^*P)Q$. This doesn't seem to get me there, for I have no way of rewriting the expression $(PQ)^*(PQ) = (Q^*P^*)(PQ)$ in a way that gets me closer to $(P^*P)(Q^*Q)$. I've tried going back to the definition of octonion multiplication and writing everything in terms of quaternions, but this gets ugly very fast and seems to require that I go even one layer deeper and write everything in terms of complex numbers, which is far too tedious. Is there a clean way to prove this identity, using just the tools I've mentioned?
Edit: Here's my currect progress: $|P|^2|Q|^2 = (P^*P)(Q^*Q) = ((P^*P)Q^*)Q = (Q^*(P^*P))Q$, where I have used one of the above identities together with the fact that $P^*P$ is real and thus commutes with everything. This gets the terms in the same order as $(Q^*P^*)(PQ)$, but the grouping is still different.
Edit 2: The technique of expressing everything in terms of quaternions relies on proving that the following expression is zero, with $p,q,r,s \in \mathbb{H}$: $$rqs^*p + (rqs^*p)^* - qs^*pr - (qs^*pr)^*.$$The noncommutativity of quaternions is an obstacle here.
Using that $(xy)^* = y^*x^*$, you get $$|xy|^2 = (xy)^*(xy) = y^*x^*xy = (x^*x)(y^*y) = |x|^2|y|^2 \,.$$
The identity $(xy)^* = y^*x^*$ can be verified directly from the definitions.
(By induction, you may assume that it holds in the quaternions, since $\mathbb{H}=\mathbb{C}\times\mathbb{C}$ with multiplication defined in the same way as for $\mathbb{O}$.)