In the paper "Lyndon Words, Free Algebras, and Shuffles" by Melancon and Reutenauer, the authors state a theorem that uses the fact that any word can be decomposed into Lyndon words (words on an ordered alphabet that are minimum among all of their rotations), but I don't understand their decomposition. From page 583, Theorem 2, part 3:
Let $w$ be any word, decomposed into Lyndon words as
$w=l_1^{i_1}\cdots l_k^{i_k} (l_j\in L, l_1>l_2>\cdots l_k)$
Then
$S_w=\frac{1}{i_1!\cdots i_k!}S_{l_1}^{i_1}\circ\cdots S_{l_k}^{i_k}$
where exponentiation means shuffle exponentiation.
Here, $L$ is the set of Lyndon words over this alphabet, $S_w$ is the series in the dual basis corresponding to the word $w$ (the theorem shows how to construct these), and $\circ$ is the shuffle product.
I understand that the last line means that $S_{l_1}^{i_1}$ means that series has the shuffle product taken with $i_1$ copies of itself. What I don't understand is how they wrote $w$ with that expression with exponents. I assume that those exponents mean shuffle exponentiation too, but most shuffle products (all nontrivial ones, really) produce linear combinations of words, not individual words. Then $w$ looks like I have to apply what I assume is concatenation (if not that, multiplication) to those linear combinations, which should also be a linear combination, not a single word. The only way this makes sense to me is if each $i_j=1$, which is trivial. Can someone explain what that expression means, and how to decompose an arbitrary word in that fashion?
The indices $i_1, \dots, i_k$ are simply positive integers and refer to the concatenation product and not to the shuffle product. It would be equivalent to write $w = l_1 \dotsm l_n$ with $l_1 \geqslant l_2 \geqslant \dotsm \geqslant l_n$.