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
$$\|{\bf A}\|^2=\sum\_{k=1}^d\|A\_k\|^2.$$


If ${\bf T}=(T\_1,...,T\_d) \in \mathcal{B}(F)^d$ and ${\bf S}=(S\_1,\cdots,S\_m)\in \mathcal{B}(F)^m$, we define $\mathbf{T}\mathbf{S}$ by
$$\mathbf{T}\mathbf{S}:=(T\_1S\_1,\cdots,T\_1S\_m,T\_2S\_1,\cdots,T\_2S\_m,\cdots,T\_dS\_1,\cdots,T\_dS\_m).$$
Let $\mathbf{T}^2=\mathbf{T}\mathbf{T}$ and $\mathbf{T}^{n+1}=\mathbf{T}\mathbf{T}^n$ 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 want to prove that
>  $$\|\mathbf{T}^n\|^2=\sum\_{g\in \mathbf{G}(n,d)}\|\mathbf{T}\_g\|^2\,$$
>  where $\mathbf{G}(n,d)$ is the set of all functions from $\{1,\cdots,n\}\to\{1,\cdots,d\}$ and $\mathbf{T}\_g:=\prod\_{i=1}^dT\_{g(i)}$.
> 
> 
> 



Proving that $\|\mathbf{T}^n\|^2=\sum_{g\in \mathbf{G}(n,d)}\|\mathbf{T}_g\|^2\,$
I believe you meant to write $\mathbf{T}\_g:=\prod\_{i=1}^n T\_{g(i)}$ so maybe a better symbol would be $\mathbf{T}\_g^n:=\prod\_{i=1}^n T\_{g(i)}$.


Now, follwing @DongryulKim's suggestion, we can see why your claim holds by observing how $\mathbf{T}^n$ is formed. 


Let us characterize functions $g \in G(n,d)$ by the values they take $g\_{i\_1, i\_2, \dots , i\_d} \in G(n,d)$, meaning $g\_{i\_1, i\_2, \dots , i\_d}(k) = i\_k$ with $k \in {1, 2, \dots , n}$ and $i\_k \in {1,2,\dots,d}$. 


Clearly, when $N=1$ the components run over all the functions in $G(1,d) = \{g\_1, g\_2 , \dots, g\_d \}$ since


$$\mathbf{T}^1 = (T\_1, T\_2, \dots , T\_d) = \left(T\_{g\_1(1)}, T\_{g\_2(1)}, \dots, T\_{g\_d(1)} \right ) = \left(\mathbf{T}\_{g\_1}^1, \mathbf{T}\_{g\_2}^1, \dots, \mathbf{T}\_{g\_d}^1 \right ) $$


Let us assume now that this also holds for $N=n$, that is
$$\mathbf{T}^n = (\mathbf{T}\_{g\_{1,1,\dots,1}}^n, \dots, \mathbf{T}\_{g\_{i\_1,i\_2,\dots,i\_n}}^n, \dots , \mathbf{T}\_{g\_{d,d,\dots,d}}^n)$$


Now, going to $N=n+1$ we have


$$\mathbf{T}^{n+1} = (T\_1 \mathbf{T}\_{g\_{1,1,\dots,1}}^n, \dots ,T\_1 \mathbf{T}\_{g\_{d,d,\dots,d}}^n ,
T\_2 \mathbf{T}\_{g\_{1,1,\dots,1}}^n, \dots, T\_2 \mathbf{T}\_{g\_{d,d,\dots,d}}^n , \cdots ,
T\_d \mathbf{T}\_{g\_{1,1,\dots,1}}^n, \dots, T\_d \mathbf{T}\_{g\_{d,d,\dots,d}}^n
 )$$


It is easy now to see from this arrangement that the first block of components corresponds to terms $\mathbf{T}\_g^{n+1}$ for functions $g \in G(n+1,d)$ with $g(1) = 1$ and $g(i) = g'(i-1)$ for any function $g' \in G(n,d)$ and $i>1$. Similarly, the second block corresponds to terms $\mathbf{T}\_g^{n+1}$ for functions $g \in G(n+1,d)$ with $g(1) = 2$ and $g(i) = g'(i-1)$ for any function $g' \in G(n,d)$ and $i>1$. Following this all the way to the last block, where $g(1) = d$ we see that we have covered all the functions in G(n+1,d). Thus


$$\mathbf{T}^{n+1} = (\mathbf{T}\_{g\_{1,1,\dots,1}}^{n+1}, \dots,\mathbf{T}\_{g\_{i\_1,i\_2,\dots,i\_d}}^{n+1},\dots,\mathbf{T}\_{g\_{d,d,\dots,d}}^{n+1}
 )$$


So, in general, we have


$$\mathbf{T}^n = (\mathbf{T}\_{g\_{1,1,\dots,1}}^n, \dots ,\mathbf{T}\_{g\_{i\_1,i\_2,\dots,i\_d}}^n,\dots,\mathbf{T}\_{g\_{d,d,\dots,d}}^n
 )$$


from which we can get


$$
\|\mathbf{T}^n\|^2=\sum\_{(i\_1,...,i\_n) \in \{1,...,d\}^n}\|\mathbf{T}\_{g\_{i\_1,...,i\_n}}^n\|^2 = 
\sum\_{g\in \mathbf{G}(n,d)}\|\mathbf{T}\_g^n\|^2
$$


