Yes it is of this form because any Lie polynomial
"begins" by a Lyndon word, in particular, the least
monomial of $B(l)$ is $l$. Then
$$
B(m)=m+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}u;\ B(n)=n+\sum_{m<v\atop |v|=|n|} \alpha^{(v)}v
$$
then $[B(m),B(n)]$ has only words of length $|m|+|n|$ and its least word is $mn$.
Computationally
- $$ B(m)B(n)=mn+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}mv+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}un]=mn+[sb1] $$ all the monomials within square brackets are of same length and strictly greater than $mn$
- $$ B(n)B(m)=nm+[\sum_{m<v\atop |v|=|n|} \alpha^{(v)}vm+\sum_{m<u\atop |u|=|m|} \alpha^{(u)}nu]=nm+[sb2] $$ all the monomials within square brackets are of same length and strictly greater than $nm$.
- But, $mn<nm$ because $mn$ is Lyndon and then $$ [B(m),B(n)]=mn+[sb1]-nm-[sb2]=mn+[sb3] $$ where the square bracket $[sb3]$ is a linear combination of monomials that are greater than $mn$ or $nm$