Show that norm of normal operator equals spectral radius via $\| T T^* \| = \| T \|^2 = \| T \|^2$ 
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 $\| T^2 \| = \| T \|^2$.
 A: Yes, your step is valid. To see it you can use a sequence, that is, if $M:=\sup_{\|x\|=1}\|Tx\|<\infty $ then there is a sequence $(x_n)$ of vectors with $\|x_n\|=1$ for all $n$ such that $\lim_{n\to \infty }\|Tx_n\|=M$. But $f(x):=x^2$ is a continuous real-valued function and so
$$
\lim_{n\to \infty }f(\|Tx_n\|)=f(\lim_{n\to \infty }\|Tx_n\|)=f(M)=M^2
$$
A: 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.
