Question about a proof involving polar decomposition of an operator Consider the following theorem in Murphy's book &quot;$C^*$-algebras and operator theory&quot;:

Can someone explain why the marked equality is correct? I know $|u^*|^2 = uu^* = w |u|^2 w^*$ but can't see why this equals $(w|u|w^*)^2$.
Context:
$\Vert \cdot \Vert _1$ is the trace class norm.
In Murphy's book theorem 2.3.3 it is proven (see the implication $(4) \implies (2)$) that if $w$ is a partial isometry, then $w^*w$ is the projection on $\ker(w)^\perp =  \ker (u)^\perp  = \ker (|u|)^\perp = \overline{\operatorname{im}(|u|^*)}= \overline{\operatorname{im}(|u|)}$. Hence, $w^*w$ is the identity on the image of $|u|$ and consequently
$$(w|u|w^*)^2 = w|u|w^*w|u|w^* = w|u|^2w^*$$
$(w\vert u \vert w^*)^2 = (w \vert u \vert w^*)(w \vert u \vert w^*) = w \vert u \vert (w^*w) \vert u \vert w^* = \ldots$