Prove that $\|\mathbf{T}^n\|^2=\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|\mathbf{T}^{\alpha}\|^2.$ Let $E$ be a  complex Hilbert space and  $\mathcal{L}(E)$ be the algebra of all bounded linear operators on $E$.
For ${\bf A} = (A_1,...,A_d) \in \mathcal{L}(E)^d$, the norm of ${\bf A}$ is given by
$$\|{\bf A}\|^2=\sum_{k=1}^d\|A_k\|^2.$$
For ${\bf T}=(T_1,...,T_d) \in \mathcal{L}(E)^d$ and ${\bf S}=(S_1,\cdots,S_m)\in \mathcal{L}(E)^m$, 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 ${\bf T}=(T_1,...,T_d) \in \mathcal{L}(E)^d$ be such that $T_iT_j=T_iT_j$ for all $i,j\in 1,\cdots,d$. I want to prove that
  $$\|\mathbf{T}^n\|^2=\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|\mathbf{T}^{\alpha}\|^2.$$
  Note that 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 $\mathbf{T}^\alpha:=T_1^{\alpha_1} \cdots T_d^{\alpha_d}$.

 A: The reason why there is a factor $\frac{n!}{\alpha!}$ is that with the multi-index notation, the order with which the operator components are composed is undistinguished. For instance, if $d=2$, then since $T_iT_j=T_jT_i$ for all $i,j$, we have
$$T_1T_2=T^{(1,1)}=T_2T_1 $$
Therefore, you have
$$\|T^n\|^2=\sum_{|\alpha|=n}c_{\alpha}\|T^{\alpha}\|^2 $$
where $c_{\alpha}$ is the number of possible permutations of a composition of $n$ components of $T=(T_1,\dots,T_d)$ which correspond to the multi-index $\alpha$. 
\begin{align*}c_{\alpha}&=\#\left\{(j_1,\dots,j_n):T_{j_1}\dots T_{j_n}=T^{\alpha},\,j_k\in \left\{0,\dots,d\right\}\right\} = \\ 
&=\#\left\{(j_1,\dots,j_n):(\#\left\{j_k=1\right\}=\alpha_1 \land \dots \land  \#\left\{j_k=d\right\}=\alpha_d)\right\}\\ 
\end{align*}
To compute the cardinality of the above set in an intuitive way, we may write out the multi-index $\alpha$ in the following way
$$\underbrace{1,\dots, 1}_{\alpha_1\text{ times }},\underbrace{2,\dots, 2}_{\alpha_2\text{ times }},\dots, \underbrace{d,\dots, d}_{\alpha_d\text{ times }} $$
Notice that the total amount of numbers written in the above line is $\alpha_1+\dots+\alpha_d=n$.
Then $c_{\alpha}$ is simply the number of possible permutations of this list, where copies of the same number are undistinguished. And this is just
$$ c_{\alpha}=\frac{n!}{\alpha_1!\cdot \dots \cdot \alpha_d!}=\frac{n!}{\alpha!}$$
