Prove that $v=vv^*v$ for a partial isometry $v$. Let $A$ be a $C^*$-algebra and let $v$ be a partial isometry so that $v^*v$ is a projection. Show that $v=vv^*v$.
The hint in the book recommended setting $z=(1-vv^*)v$ and then computing $z^*z$. I recognise that we are likely trying to show $z^*z=0$ to prove that $z=0$, however I am not sure how to show this.
edit: Using a suggestion bellow I was able to get this
\begin{align*}
    z^*z&=v^*(1-vv^*)^*(1-vv^*)v\\
    &=v^*(1-v^*v)(1-vv^*)v\\
    &=v^*(1-vv^*-v^*v+v^*vvv^*)v\\
    &=(v^*-v^*vv^*-v^*v^*v+v^*v^*vvv^*)v\\
    &=v^*v-v^*vv^*v-v^*v^*vv+v^*v^*vvv^*v\\
    &=v^*v-(v^*v)^2-v^*v^*vv+v^*v^*vvv^*v\\
    &=-v^*v^*vv+v^*v^*vvv^*v\\
    &=v^*v^*v(vv^*v-v)
\end{align*}
But I am unsure of how to proceed.
 A: In your attempt you are commuting $v$ and $v^*$ for no reason; you can't do that.
You know that $v^*v$ is a projection. That says that $v^*vv^*v=v^*v$, which we may write as $$\tag1v^*(1-vv^*)v=0.$$
Since $v^*v$ is a projection, we have $1=\|v^*v\|=\|vv^*\|$. Then $1-vv^*\geq0$. So there exists a positive square root $y$ with $y^2=1-vv^*$. With this,  $(1)$ becomes $$ v^*y^*yv=0, $$ and so $yv=0$. Then $(1-vv^*)v=y^2v=0$.
A: Here is another answer. We can get that $vv^*$, hence $1 - vv^*$, is a projection, without alluding to the existence of square roots for positive operators. For any unital Banach algebra and any elements $a,b$ in it, $\sigma(ab) \cup \{0\} = \sigma(ba) \cup \{0\}$ (so this will hold in any C*-algebra since it will hold in the unitization of it). Since $v^*v$ is a projection and therefore has spectrum contained in $\{0,1\}$, $\sigma(vv^*) \subseteq \{0,1\}$. But this implies that $(vv^*)^2 = vv^*$, so that its idempotent. (one can use the spectral mapping theorem for polynomials, along with the fact that the spectral radius of a self-adjoint is equal to its norm, to conclude that norm of $(vv^*)^2 - vv^*$ is 0).
Its also clearly self-adjoint, so its a projection. This gives that $1 - vv^*$ is a projection, so now we go about it the way you were thinking:
$$ \|((1 - vv^*)v)^*(1 - vv^*)v\| = \|v^*(1 - vv^*)^*(1 - vv^*)v\| = \|v^*(1 - vv^*)v\|,$$
which is 0, as in Martin's answer, since $v^*v$ is a projection. So this implies that $(1 - vv^*)v = 0$ using the C*-equality.
