Spectral radius of a normal operator

Question

I am stuck in intermediate step of a proof.

Let $||A^2|| = ||A||^2$ = $||A^*A||$

Then how to show this?

$||{A^2}^n||^2$ = $||{{A^*}^2}^n {A^2}^n||$ = ${{||A^*A||}^2}^n$.

Here $A^*$ denotes the Hilbert-Adjoint Operator.

Answer 1

Let's call 1) $\|A^2\|=\|A\|^2$, and 2) $\|A\|^2=\|A^*A\|$.

Then $\|A^{2^n}\|^2=\|(A^{2^n})^*A^{2^n}\|$ by 2).

$\|(A^{2^n})^*A^{2^n}\|=\|A^{*^{2^n}}A^{2^n}\|$, since $(A^k)^*=(A^*)^k$ for every $k$.

$\|A^{*^{2^n}}A^{2^n}\|=\|(A^{*}A)^{2^n}\|=\|\big((A^{*}A)^{2^{n-1}}\big)^2\|=\|(A^{*}A)^{2^{n-1}}\|^2$ by 1) and $A$ being normal. Now you proceed by induction until you get to $\|A^{*^{2^n}}A^{2^n}\|=\|A^*A\|^{2^n}$.