In this question $E$ stands for a complex Hilbert space and let $\mathcal{L}(E)$ the algebra of all bounded linear operators from $E$ to $E$.

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
$$\displaystyle\sup_{\|x\|=1}\displaystyle\sum_{|\alpha|=n}\frac{n!}{\alpha!}\|{\bf S}^{\alpha}x\|^2=||{\bf S}^n||^2,$$
with $n\in\mathbb{N}^*,\;$ $\alpha = (\alpha_1, \alpha_2,...,\alpha_d) \in \mathbb{Z}_+^d;\;\alpha!: =\alpha_1!...\alpha_d!,\;|\alpha|:=\displaystyle\sum_{j=1}^d|\alpha_j|$;  ${\bf S}^\alpha:=S_1^{\alpha_1} \cdots S_d^{\alpha_d}$ and ${\bf S}^n:={\bf S}\star{\bf S}\star\star\star{\bf S}$ with ${\bf S}\star{\bf S}$ is defined as:
${\bf S}^2 :={\bf S}\star{\bf S}= (S_1 S_1,\cdots,S_1 S_d,S_2S_1,\cdots,S_2S_d,S_dS_1\cdots,S_d S_d)$