Product of partial isometry and isometry is a partial isometry

This is an exercise from Wegge Olsen's K-theory and C*-algebras. In the exercise 5.A (b) he states that the product of an isometry by a partial isometry in a C*-algebra is a partial isometry. Taking $u$ a partial isometry (this means $uu^*u=u$) and $v$ isometry (this means $v^*v=1$) I managed to show that $vu$ is a partial isometry. I also tried to show that $uv$ is a partial isometry without success, is this even true? Doing some research, I found this question that states that $vv^*$ and $u^*u$ should commute for this to happen, I don't think having $v$ be an isometry is a sufficient condition for this.

There's a big chance that the book is only asking me to show that $vu$ is the partial isometry but I'd like to confirm that.

As you noticed, the book is only talking about $vu$.
As an example, consider first, in $M_2(\mathbb C)$, $$u=\begin{bmatrix}1/2&1/2\\1/2&1/2\end{bmatrix},\ \ \ v=\begin{bmatrix}0&0\\1&0\end{bmatrix}.$$ Here both $u,v$ are partial isometries, but $$uv=\begin{bmatrix} 1/2&0\\1/2&0\end{bmatrix}$$ is not. This behaviour can be pushed to infinite dimension: on $\ell^2(\mathbb N)$, let $S$ the unilateral shift, and $U$ the orthogonal projection onto the span of $e_1+e_2$. Then the matrix of $U$ is that of $u$ and the rest of the entries are zeros, and $$US=\begin{bmatrix} 1/2&0&0&\cdots \\ 1/2&0&0&\cdots\\ 0&0&0&\cdots\\ \vdots&\vdots&\ddots \end{bmatrix},$$ which is not a partial isometry, while $U$ is a partial isometry (a projection, in fact) and $S$ is an isometry.