Step in proof that $[A,B]=\langle C^{AB}\rangle$

Question

Let $A,B$ be subgroups of $G$. Let $\{a_i\}$, $\{b_i\}$ be generators of $A$ resp. $B$. Define $$C=[\{a_i\},\{b_i\}]=\{[a_i,b_j]\}$$ We want to prove that $$[A,B]=\langle C^{AB}\rangle $$ To prove the inclusion $[A,B]\subset \langle C^{AB}\rangle$ the author starts by noting that $$(a_{i_1}\dots a_{i_n})^{b_j}=(a_{i_1}[a_{i_1},b_j])\dots(a_{i_n}[a_{i_n},b_j])$$ which is clear. However, the author then writes $$(a_{i_1}[a_{i_1},b_j])\dots(a_{i_n}[a_{i_n},b_j])=a_{i_1}\dots a_{i_n}f$$ for some $f\in \langle C^{A}\rangle$. This step I do not understand. If I write it out of the case $n=2$: $$a_1[a_1,b]a_2[a_2,b]=a_1a_1^{-1}b^{-1}a_1ba_2a_2^{-1}b^{-1}a_2b$$ then I just don't see how to group the terms to put it in the form $a_1a_2f$ for some $f\in \langle C^A\rangle$.

Answer 1

$$a_1[a_1,b]a_2[a_2,b]=a_1a_2[a_1,b][[a_1,b],a_2][a_2,b]=a_1a_2[a_1,b]^{a_2}[a_2,b]$$