Q:

If $A$ is a self-adjoint operator .  Is $\|A^n\|=\|A\|^n,\;\forall n$? 

Let $E$ be an infinite-dimensional complex Hilbert space. 

Let $A\in \mathcal{L}(E)$ be a self-adjoint operator i.e. $A^*=A$, is
  $$\|A^n\|=\|A\|^n,\;\forall n\in \mathbb{N}?$$

Note that if $A$ is self-adjoint, then
$$\|A\|=\sup\left\{|\langle Ax, x\rangle|\,;\;x\in E,\;\|x\|= 1\right\}.$$
It is well known that if $T\in \mathcal{L}(E)$, then
$$\|T^*T\|=\|T\|^2.$$
So since $A$ is self-adjoint, then
$$\|A^2\|=\|A\|^2.$$
I try to show the result by induction. 

A:

It is known that all elements in the spectrum of a selfadjoint operator are approximate eigenvalues.  So there exists a sequence $\{x_j\}$ with $\|x_j\|=1$ for all $j$ and $Ax_j-\|A\|\,x_j\to0$. 
Now we proceed by induction on $n$. If $A^nx_j-\|A\|^nx_j\to0$, then 
$$
A^{n+1}x_j-\|A\|^{n+1}x_j=A(A^nx_j-\|A\|^nx_j)+\|A\|^n(Ax_j-\|A\|x_j)\to0. 
$$
It follows that, for each $n$, $A^nx_n-\|A^n\|x_j\to0$. This implies that $A^n-\|A\|^nI$ is not invertible, that is $\|A\|^n\in\sigma(A^n)$. But then
$$
\|A\|^n\leq\|A^n\|,
$$
which is the non-trivial inequality.  Thus, $\|A^n\|=\|A\|^n$ for all $n$. 

