Let $H$ be a Hilbert spaces and $T \in L(H)$ normal, i.e. $T T^* = T^* T$. Show that $r(T) = \| T \|$, (where $r(T) := \sup_{\lambda \in \sigma(T)} | \lambda |$ is the spectral radius of $T$) by first showing $$\| T^* T \| = \| T^* T \| = \| T \|^2 = \| T^2 \|.$$
I know that $r(T) = \lim_{n \to \infty} \| T^n \|^{\frac{1}{n}}$, but it suffices to look at that limit for $n = 2^k$. So once I have $\| T^2 \| = \| T \|^2$ I should get $\| T^{2^k} \| = \| T \|^{2^k}$ and therefore $$ r(T) = \lim_{n \to \infty} \| T^n \|^{\frac{1}{n}} = \lim_{k \to \infty} \| T^{2^k} \|^{\frac{1}{2^k}} = \lim_{k \to \infty} \| T \|^{\frac{2^k}{2^k}} = \| T \|. $$ This can be done analogously for $\| T T^* \| = \| T \|^2$.
I also know that as $A := T^* T$ is self-adjoint we have $\| A \| = \sup_{\| x \| = 1} \langle A x, x \rangle$ and therefore $$ \| T^* T \| = \sup_{\| x \| = 1} \langle T^* T x, x \rangle = \sup_{\| x \| = 1} \| T x \|^2 \overset{(\star)}{=} \left(\sup_{\| x \| = 1} \| T x \|\right)^2 = \| T \|^2, $$ but I am unsure about the step $(\star)$, is it valid?
Furthermore I am looking for a hint to show that $\| T^2 \| = \| T \|^2$.
Continuing from my impartial solution above using @DisintegratingByParts help:
We have $$ \| T^2 x \|^2 = \langle T^* T^* T T x, x \rangle = \langle T^* T T^* T x, x \rangle = \langle T^* T x, T^* T x \rangle = \| T T^* x \|^2. $$ By taking the square root and $\sup_{\| x \| = 1}$, we can conclude $\| T^2 \| = \| T T^* \| = \| T \|^2$.
For powers of two we can inductively conclude $\| T^{2^k} \| = \| T \|^{2^k}$ for $k \in \mathbb N$ as then we can always use the normality property $T T^* = T^* T$ pairwisely, therefore $$ r(T):= \lim_{n \to \infty} \| T^n \|^{\frac{1}{n}} = \lim_{k \to \infty} \| T^{2^k} \|^{\frac{1}{2^k}} = \lim_{k \to \infty} \| T \|^{\frac{2^k}{2^k}} = \| T \|, $$ holds, because if the limit of a sequence exists, then the limit of every subsequence is the equal to that limit.