Normal Operators (Obsolete!) Problem
Given a Hilbert space $\mathcal{H}$.
Consider a normal operator:
$$N:\mathcal{D}(T)\to\mathcal{H}:\quad N^*N=NN^*$$
Construct the operator:
$$A:=N(1+N^*N)^{-1}\in\mathcal{B}(\mathcal{H})$$
Then it is normal:
$$A^*\in\mathcal{B}(\mathcal{H}):\quad A^*A=AA^*$$
How can I prove this?
Disclaimer
This question became obsolete due to: Transform
I kept it open for the sake of respect for TAE's answer.
Also it may serve as a manual to handling such problems.

By the other thread:
$$WN\subseteq NW\quad WN^*\subseteq N^*W$$
And formally it holds:
$$\psi\in\mathcal{D}(N^*):\quad A^*\psi=WN^*\psi$$
So a calculation gives:
$$AA^*\psi=NWWN^*\psi=WNN^*W\psi=WN^*NW\psi=A^*A\psi\quad(\psi\in\mathcal{D}N^*)$$
But the domain was dense:
$$\overline{\mathcal{D}(N^*)}=\mathcal{H}\implies AA^*=A^*A$$
That finishes the proof.
