Consider a binary shuffle algebra $\mathcal{W}$ of two letters $a, b$. As usual the concatination of two words $u = u_1 \dots u_m$, $v = v_1 \dots v_n$ is defined as: $$u \bullet v := u_1 \dots u_m v_1 \dots v_n$$ and the shuffle product is defined recursively as: $$(k \bullet u) * (l \bullet v) := k \bullet (u * (l \bullet v) ) + l \bullet ( (k \bullet u) * v)$$ where $k,l \in \{a,b\}$ are some letters... If necessary, the coproduct of $w \in \mathcal{W}$ is: $$\Delta (w) = \sum_{u \bullet v = w} u \otimes v.$$
An element $c \in \mathcal{W}$ is said to be in echalon form of weight $N$, if it is the concatination exponent of some linear combination of $a$ and $b$, i.e.: $$c = (\alpha a + \beta b)^N = \sum_{k=0}^N \alpha^{N-k} \beta^{k} (a^{N-k} * b^{k}).$$
The problem I have is to prove (relatively easy) that any shuffle element $a^m * b^n$ can be represented as a combination of echalon elements of wieght $m+n$ and to find a formula (the hard part) for this, e.g. something like: $$a^m * b^n = \sum_{k=1}^{m+n} \gamma_k^{(m,n)} (\alpha_k^{(m,n)} a + \beta_k^{(m,n)} b)^{m+n} \, ?$$ Examples: $$a*b = ab + ba = \frac{1}{2} \left( (a+b)^2 - (a-b)^2 \right),$$ $$a^2 * b = a^2b + aba + ba^2 = \frac{1}{2} \left( (a+b)^3 - (a-b)^3 - 2 b^3 \right)...$$
I've found by applying the difference formula:
$$\Delta^n f(x) = \sum_{k=0}^n (-1)^{n-k} \binom{n}{k} f(x + k),$$
where $\Delta f(x) = f(x+1) - f(x)$, on $f(x) = x^{n+1}$:
$$ \frac{n}{2} + x = \sum_{k=0}^n \frac{(-1)^{n-k}}{(n+1)!} \binom{n}{k} (x + k)^{n+1}, \\ \Rightarrow a * b^n = \sum_{k=0}^n \frac{(-1)^{n-k}}{n!} \binom{n}{k} (a + k \, b)^{n+1} - \binom{n+1}{2} b^{n+1},$$
which provides a recursive step that solves the given problem.