The lower $p$-series of a group $G$ is defined to be the descending series $$G=P_{1}(G) \geq P_{2}(G) \geq \dots, \; \text{where $P_{i+1}(G) = P_{i}(G)^{p}[P_{i}(G),G]$}.$$
I'm looking for a proof that this series is strongly central, i.e that $[P_{i}(G), P_{j}(G)] \subseteq P_{i+j}(G)$ for all$ i, j \in \mathbb{N}$.
I'm currently reading through a paper on $p$-adic analytic groups, and immediately after defining the lower $p$-series, the author simply states that "this is a basic property of the series" without proof.
It seems sensible to proceed with a proof by induction since both $i$ and $j$ are natural numbers. My attempts so far have been unsuccessful; I'm having trouble using the inductive hypothesis in any useful way. Having checked through some references, it would appear that a proof for many basic facts about the lower $p$-series are proven in "Finite Groups II" by Huppert and Blackburn, but I don't have access to the book. Any help would be appreciated.
We want to prove $[P_i(G),P_j(G)] \le P_{i+j}(G)$. We do it by induction on $i+j$ and then on $i$, where we can assume that $i \le j$. It's true for $i=1$ by definition. So assume $i,j > 1$. I will just write $P_i$ for $P_i(G)$.
Then $$[P_i,P_j] = [[G,P_{i-1}]P_{i-1}^p,[G,P_{j-1}]P_{j-1}^p]] \le ABC,$$ where $A=[[G,P_{i-1}],[G,P_{j-1}]]$, $B=[P_{i-1}^p,P_j]$, and $C=[P_i,P_{j-1}^p]$.
Now by the $3$-subgroups lemma, we have $$A \le [[G,[G,P_{j-1}]],P_{i-1}][G,[[G,P_{j-1}],P_{i-1}]]\le [P_{j+1},P_{i-1}][G,P_{i+j-1}] \le P_{i+1},$$ by induction.
Now, by induction, we have $[P_{i-1},P_j] \le P_{i+j-1}$ and then, since $P_{i+j-1}/P_{i+j} \le Z(G/P_{i+j})$, we have $$B=[P_{i-1}^p,P_j] \le P_{i+j-1}^pP_{i+j} \le P_{i+j}.$$
The proof for $C$ is similar.