Let ${\bf S} = (S\_1,...,S\_d) \in \mathcal{L}(E)^d$. We recall that the norm of $\|{\bf S}\|$ is defined by
\begin{eqnarray\*}
\|{\bf S}\|
&=&\sup\left\{\bigg(\displaystyle\sum\_{k=1}^d\|S\_kx\|^2\bigg)^{\frac{1}{2}},\;x\in E,\;\|x\|=1\;\right\},
\end{eqnarray\*}


I want to show that if the operators $S\_k$ are commuting, then
$$\displaystyle\sup\_{\|x\|=1}\displaystyle\sum\_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2,$$
where ${\bf S}^n:={\bf S}\cdot{\bf S}\cdot\cdots\cdot{\bf S}$.


Note that ${\bf S}^2 :={\bf S}\cdot{\bf S}= (S\_1 S\_1,\cdots,S\_1 S\_d,S\_2S\_1,\cdots,S\_2S\_d,S\_dS\_1\cdots,S\_d S\_d)$, and ${\bf S}^n$ is defined by induction.


Thank you!



Why this equality holds?
As it was mentioned in the comments, this is basically writing down the definition.


Since all the $S\_i$'s commute with each other, then each component of $\boldsymbol{S}^n$ is of the form $(S\_1 \ldots S\_1)(S\_2 \ldots S\_2) \ldots (S\_d \ldots S\_d)$, where $S\_i$ apprears $\alpha \_i$ times and $\sum \_i \alpha \_i =n$. The number of times the sequence $\alpha = (\alpha \_1 , \ldots , \alpha \_d)$ apprears, is the number of ways you can choose the first option(which is $S\_1$) $\alpha\_1$ times, the second option (that is $S\_2$) $\alpha\_2$ times and so on, and this is given by the binomial expansion ${n \choose{\alpha\_1, \alpha\_2, \ldots , \alpha\_d }} = \frac{n!}{\alpha\_1!\ldots \alpha\_d!}$. 


Hence, we get:
$$||S^n||^2=\sup\_{||x||=1} \sum\_{|\alpha|=n}\frac{n!}{\alpha!}||S^{\alpha}x||^2,$$


because the number of $ S\_{i\_1}S\_{i\_2}...S\_{i\_n}$ in the $d^n$ tuple which will become $S\_1^{\alpha\_1}...S\_d^{\alpha\_d}$ is $\frac{n!}{\alpha!}$, where each $i\_j\in \{1,2,...,d\}$.


