Question about a proof involving polar decomposition of an operator

Question

Consider the following theorem in Murphy's book "$C^*$-algebras and operator theory":

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.

Answer 1

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 $$(w|u|w^*)^2 = w|u|w^*w|u|w^* = w|u|^2w^*$$

Answer 2

$(w\vert u \vert w^*)^2 = (w \vert u \vert w^*)(w \vert u \vert w^*) = w \vert u \vert (w^*w) \vert u \vert w^* = \ldots$