Our proof is a little bit different than the proofs I've found here on this forum. I understand the rest of the proof as it's just using the Gelfand-Beurling formula, but the part I don't understand goes like this (that's the first part of the induction to show that $\|A\|^{2^k} = \|A^{2^k}\|$):
$$\|A^4\| = \|A^* A\|^2 = \|(A^* A)^* \ (A^* A)\| = \|(A^*A)(A^*A)\| = \|(A^*)^2 (A)^2\| = \|A^2\|^2$$
I basically don't understand every " $=$" here. Could someone maybe help me understand this part of the proof step by step?
I think it could maybe have something to do with the fact, that $\|A^* A\| = \|A\|^2$, that multiplying operators could here be commutative and that maybe for some reasons it's true here that $\|A^k \| = \|A\|^k$ for $k$ natural. But still, even assuming those 3 things are true and used here, I don't see where the third "$=$" is coming from.
The only things one uses are the C$^*$-identity: $$\tag1\|T\|^2=\|T^*T\|,$$ and that $A^*A=AA^*$, which is normality. That said, I don't understand the steps as written in your question. Here is my take.
You start with $\|A^4\|=\|A^{*4}A^4\|^{1/2}$ by applying $(1)$.
By normality this is equal to $\|(A^*A)^4\|^{1/2}$.
Now we use that $A^*A$ and its square are selfadjoint. Then, using $(1)$ in the second equality, $$\|(A^*A)^4\|^{1/2}=\|(A^*A)^{2*}(A^*A)^2\|^{1/2}=\big[\|(A^*A)^2\|^2\big]^{1/2}=\|(A^*A)^2\|. $$
Using that $A^*A$ is selfadjoint and using $(1)$ twice, we get $$ \|(A^*A)^2\|=\|(A^*A)^*(A^*A)\|=\|A^*A\|^2=\big(\|A\|^2\big)^2=\|A\|^4. $$