Let $F$ be a complex Hilbert space and $\mathcal{B}(F)$ be the algebra of all bounded linear operators on $F$.

For ${\bf A} = (A_1,...,A_d) \in \mathcal{B}(F)^d$, the norm of ${\bf A}$ is given by $$\|{\bf A}\|^2=\sum_{k=1}^d\|A_k\|^2.$$

For ${\bf T}=(T_1,...,T_d) \in \mathcal{B}(F)^d$ and ${\bf S}=(S_1,\cdots,S_m)\in \mathcal{B}(F)^m$, we set $$\mathbf{T}\mathbf{S}:=(T_1S_1,\cdots,T_1S_m,T_2S_1,\cdots,T_2S_m,\cdots,T_dS_1,\cdots,T_dS_m).$$ Let $\mathbf{T}^2=\mathbf{T}\mathbf{T}$ and we define by induction $\mathbf{T}^{n+1}=\mathbf{T}\mathbf{T}^n$ for $n\in \mathbb{N}^*$.

Let $n\in \mathbb{N}^*$ and ${\bf T}=(T_1,...,T_d) \in \mathcal{B}(F)^d$ such that the operators $T_k$ are not necessarily commuting, I

claimthat $$\|\mathbf{T}^n\|^2=\sum_{g\in \mathbf{G}(n,d)}\|\mathbf{T}_g\|^2\,$$ where $g\in \mathbf{G}(n,d)$ and $\mathbf{T}_g:=\prod_{i=1}^dT_{g(i)}$.Note that $\mathbf{G}(n,d)$ denotes the set of all functions from $\{1,\cdots,n\}$ into $\{1,\cdots,d\}$.

How to prove that the claim is true?

Note that if the operators $T_k$ are commuting, it can be seen that $$\|\mathbf{T}^n\|^2=\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|\mathbf{T}^{\alpha}\|^2.$$ For $\alpha = (\alpha_1,\cdots,\alpha_d) \in \mathbb{N}^d$, we write $\alpha!: =\alpha_1!\cdots\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$ and $\mathbf{T}^\alpha:=T_1^{\alpha_1} \cdots T_d^{\alpha_d}$.