Campbell-Baker-Hausdorff Theorem Proof from Stillwell I am currently working through the proof of the Campbell-Baker-Hausdorff theorem in Stillwell's Naive Lie Theory. He starts with letting 
$$e^Ae^B=e^Z, \qquad Z=\sum_{i=1}^\infty F_i(A,B)\qquad (*)$$
where $F_n(A,B)$ is the sum of all the terms of degree $n$ in $Z$, and hence a homogeonous polynomial of degree $n$ in the variables $A$ and $B$. He goes on to say that we will call a polynomial $p(A,B,C,\dots)$ Lie if it is a linear combination of $A,B,C,\dots$ and (possible nested) Lie bracket terms in $A,B,C,\dots$ . 
The Campbell-Baker-Hausdorff theorem then becomes: For each $n\geq 1$, the polynomial $F_n(A,B)$ in $(*)$ is Lie.
He starts the proof by writing for any $A,B,C$ we have
$$(e^Ae^B)e^C=e^A(e^Be^C)$$
and therefore if $e^Ae^Be^C=e^W$,
$$W=\sum_{i=1}^\infty F_i(\sum_{j=1}^\infty F_j(A,B),C)=\sum_{i=1}^\infty F_i(A,\sum_{j=1}^\infty F_j(B,C)).$$
We then let $n>2$ and suppose $F_m$ is a Lie polynomial for $m<n$ with the goal to show $F_n$ is Lie, which would complete our argument by induction.
He then says that the induction hypothesis implies that all homogeneous terms of degree less than $n$ in both expressions for $W$ are Lie, and so too are the homogeneous terms of degree $n$ resulting from $i>1$ and $j>1$. The only possible exceptions are
$$F_n(A,B) +F_n(A+B,C) \quad \text{on the left (from $i=1,j=n$ and $i=n,j=1$)}$$
$$F_n(A,B+C) + F_n(B,C) \quad \text{on the right (from $i=n,j=1$ and $i=1,j=n$}.$$
My question is why are the homogeneous terms of degree $n$ resulting from $i>1$ and $j>1$ also Lie with the only exceptions being the listed ones? I am also having a hard time seeing where the exceptional polynomials come from, so where should I proceed to see this?
 A: This answer only tried to give a picture vision of the situation, hoping that things become clear already. After the answer there is a cited reference, Free Lie Algebras, which is the better answer. (Because it is a structural answer, and the structure is beautiful, just switch to the reference and enjoy!)
First, in my pictural opinion a Lie polynomial in the alphabet with letters $A,B,C,D,\dots$ is a homogeneous polynomial in the non-commutative algebra generated by the monoid generated by these letters, we will work over $\Bbb Q$, that can be obtained in the following way.
First fix some letters (with possible repetitions) (from the alphabet) and some order, and put them in a row. For instance;
A B A C A D B A

Now decide to build a "special" tree with these nodes as leaves, going "down", so decide which two neighbor letters should be Lie-condensed first, then use this as a "new letter", and go on recursively. One picture may be:
A   B   A  C   A   D   B   A
 \   \ /    \   \ /   /   /
  \   *      \   *   /   /
   \ /        \ /   /   /
    *          *   /   /
     \          \ /   /
      \          *   /
       \          \ /
        \          *
         \        /
          \      /
           \    /
            \  /
             \/
            FINAL RESULT

Each * means to get the joined nodes, and apply [ , ] on them.
I hope it is clear.
Now observe that 
$$
\begin{aligned}
Z 
&= F(A,B)
\\
&=\log(e^Ae^B)
\\
&=\log\left(\ 
\left(1+\frac 1{1!}A+\frac 1{2!}A^2+\dots\right)
\left(1+\frac 1{1!}B+\frac 1{2!}B^2+\dots\right)
\ \right)
\\
&=\log\left(\ 
1+\sum_{(j,k)\ne (0,0)}
\frac 1{j!k!}A^jB^k
\ \right)
\\
&=
0+\underbrace{(A+B)}_{F_1(A,B)}+\dots
\end{aligned}
$$
has the $F_1$-part equal to $A+B$, a Lie polynomial, and the further homogeneous pieces are under attack.
Back to the question. Why is $F_i\left(A,\sum_j F_j(B,C)\right)$ inductively a Lie polynomial (for $i,j>1$)? Use new letters $D_j$ instead of $F_j(B,C)$ if this makes the things simpler, and let us make the picture of $F_i(A, \sum _j D_j)$. There are many terms that are involving tree collapsing rules as above in a linear combination. Now push each $\sum D_j$ from the sum down on its piece from $F_i$, till it hits the *, i.e. it is involved in building a Lie bracket. This bracket is linear, so we split the sum $\sum D_j$ into pieces, and work now with an individual $D_j$.
If this $D_j$ is itself (inductively) given by such Lie bracket tree collapsing rules, then we are fine, formally we "move the rule to the top.
We only have problems with $F_1$, which is not really in the range of the  Lie bracket tree collapsing rules. I cannot say more.

(I could not figure out which / where is the problem with the "exceptional polynomials", since working only with the homogeneous part of degree $(n+1)$, for instance for $i=1$, $j=n$, and conversely, LHS,
$$
\begin{aligned}
F_1(F_n(A,B),C)
&=F_n(A,B)+C\ ,
\\
F_n(F_1(A,B),C)
&=F_n(A+B,C)\ ,
\end{aligned}
$$
and of course, now we have to start the proof.)

I am now saying some words about the hidden structure, it is a wonderful structure, enjoy it!
In the book version of Free Lie Algebra, Christophe Reutenauer, 
referenced fully also in Free Lie Algebras, wiki page the author is quickly introducing a structure of a Hopf algebra on $\Bbb Q\langle\langle A, B,\dots\rangle\rangle$, the free algebra on the monoid generated by the letters $A,B,\dots$, the one multiplication is the usual one, an other one is given by the shuffle product, so for instance $A \sqcup\!\!\!\sqcup B = AB-BA=[A,B]$, so shuffle product monomials are... Lie polynomials. There are two corresponding comultiplications, and using these constructions one can state structural properties. In the list of them, relevant for the present question:


*

*Theorem 1.4 in the book, (not in the linked pdf,) characterizes a polynomial $P$ to be Lie polynomial in the following way, there are equivalent:


*

*$P$ is a Lie polynomial,

*Define $ad(P)$ by $ad(P)(Q)=[P,Q]=PQ-QP$. One can now take this setting only for generators $A,B,\dots$ of the alphabet, so $ad(A)=[A,-]$, and extend to an algebra mapping, so for instance $Ad(AB)=Ad(A)Ad(B):=ad(A)ad(B)$. The equivalent condition for a $P$ is then $ad(P)=Ad(P)$.

*$P$ is primitive, a structural property in a Hopf algebra.

*$P$ has no free coefficient and the derivative of $P$ conincides with the "right bracketing" of $P$.


*Theorem 3.1. is a version of the above for Lie series.

*Lemma 1.7 in the book, let $\alpha $ be the antipode, mapping a word $w$ into $\pm$ the reversed word, the sign being captured from the parity of the length. Then for a Lie polynomial $P$ we have $\alpha(P)=-P$.

*Theorem 3.2 in the book, let $S=1+\dots$ be a series, higher terms omitted, then there are equivalent:


*

*$\log(S)$ is a Lie series,

*$S$ is group-like, i.e. $\delta(S)=S\otimes S$,

*the map $w\to (S,w)$ is a homomorphism from the shuffle algebra to $\Bbb Q$,

*$Ad(S)(T)=STS^{-1}$.



Corollary 3.3, the series $S=1+\dots$, such that $\log S$ is Lie series, are building a group under multiplication, this is because of the group-like-property.
Corollary 3.4, $\log(e^Ae^B)$ is a Lie series. Because of the stability w.r.t. the multiplication above, and note that $\log e^A=A$, $\log e^B=B$, are Lie series.
