Is $\Vert A^n \Vert = \Vert A \Vert^n$ for normal operator $A$ on inner product space? Exercise 7(a) after $\S$ 87. Norm from Paul R. Halmos's "Finite-Dimensional Vector Spaces" (second edition) invites a comment on the following assertion.

If (linear operator) $A$ is normal, then $\Vert A^n \Vert = \Vert A \Vert^n$ for every positive integer $n$.

For reference, $\S \ 87.$ Norm (from the book) has the following definition for the norm $\Vert \cdot \Vert$ of a linear operator:

$\Vert A \Vert = \inf \big\{K: \Vert Ax \Vert \leq K \Vert x \Vert \text{ for all vectors } x \big\}.$

Going by the discussion preceding this exercise (in the book), I assume that the assertion concerns inner product spaces, not the more general normed vector spaces. However, the inner product space, say $\mathcal V$, of the assertion is not specified to be over the complex (or real) field, and is not said to be finite-dimensional. Also, $\mathcal V$ is not given to be complete.
I am able to see why the assertion holds if $\mathcal V$ is finite-dimensional over the complex field. The reason is the spectral theorem for normal operators on such vector spaces. Similarly, I also understand that the assertion would hold in another case: if $A$ is self-adjoint and if $\mathcal V$ is finite-dimensional over the real field. I am not able to imagine what happens in the general case however, and would appreciate a pointer. Thanks.
 A: I will use the equivalent definition (§88) that $\|A\|=\sup_{\|x\|=1}\|Ax\|$. I will first prove by a Cauchy-styled mathematical induction that $\|P^n\|=\|P\|^n$ for a positive operator $P$. In this case $\|P\|=\sup_{\|x\|=1}(Px,x)$ (see §89). Pick an integer $k$ such that $2^k\ge n$. Then
$$
\|P^{2^k}\|
=\sup_{\|x\|=1}(P^{2^k}x,x)
=\sup_{\|x\|=1}(P^{2^{k-1}}x,P^{2^{k-1}}x)
=\sup_{\|x\|=1}\|P^{2^{k-1}}x\|^2
=\|P^{2^{k-1}}\|^2
$$
and inductively we obtain $\|P^{2^k}\|=\|P\|^{2^k}$ and in turn $\|P\|^{2^k}=\|P^{2^k}\|\le\|P^n\|\|P\|^{2^k-n}$ by submultiplicativity. Therefore $\|P\|^n\le\|P^n\|$. Yet, by submultiplicativity, we also have $\|P^n\|\le\|P\|^n$. Hence $\|P^n\|=\|P\|^n$.
Now suppose $A$ is normal. Let $P=A^\ast A$. Then
$$
\|A^n\|
=\sup_{\|x\|=1}\|A^nx\|
=\sup_{\|x\|=1}(A^nx,A^nx)^{1/2}
=\sup_{\|x\|=1}(P^nx,x)^{1/2}
=\|P^n\|^{1/2}
$$
for every positive integer $n$, where the normality of $A$ and the positivity of $P$ are used in the third and the fourth equalities respectively. Put $n=1$, we get $\|A\|=\|P\|^{1/2}$. Hence
$$
\|A^n\|=\|P^n\|^{1/2}=\|P\|^{n/2}=\|A\|^n.
$$
