Let $E$ be a complex Hilbert space and $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For $A=(A_1,...,A_n) \in \mathcal{L}(E)^n$ and $m,k\in \mathbb{N}^*$. I want to show that $$\sum_{h\in F(m,n)} A_h^*\left(\sum_{g\in F(k,n)} A_g^* A_{g}\right) A_{h}=\sum_{f\in F(m+k,n)} A_f^* A_{f}\; ??$$ where $F(m,n)$ the set of all functions $f$ from $\{1,2, ... , m\} $ to the set $\{1.2...,n\}$. For $g\in F(m,n)$, we set $A_g:=A_{g(1)}\cdots A_{g(m)}$.
I try as follows:
$$ \sum_{h\in F(m,n)} A_h^*\left(\sum_{g\in F(k,n)} A_g^* A_{g}\right) A_{h}=\sum_{h \in F(m,n), g \in F(k,n)} (A_{g\ast h})^* A_{g \ast h}, $$
where $g \ast h$ is a function from $ F(m,n)\times F(k,n)$ to $F(m+k,n)$. I'm facing difficulties to give the expression of $g \ast h$ in order to show that it is a bijection.
What you want is an operation $\star$ such that for $f=g\star h$, $A_{f}$ is $A_g A_h=A_{g(1)}\cdots A_{g(m)}A_{h(1)}\cdots A_{h(k)}$. This means that
$$ \begin{array}{l} f(1)=g(1), \cdots,f(m)=g(m),\\ f(m+1)=h(1), \cdots,f(m+k)=h(k), \end{array}\tag{1} $$
And indeed, (1) does define a bijection from $ F(m,n)\times F(k,n)$ to $F(m+k,n)$ (it is just a concatenation when you think about it).