Let $A\in{\mathbb C}^{n,n}$ be a $n\times n$ matrix and $\omega(A) := \sup\left\{|x^HAx|:\|x\|_2=1,\,x\in \mathbb C^n\right\}$ the numerical range of $A$.
There is an elementary proof by Carl Pearcy$^{[1]}$ that if $\omega(A)\leq 1$ then follows that $\omega(A^n) \leq [\omega(A)]^n$.
He states that by the property $\omega(\lambda A) = |\lambda|\omega(A)$ it suffices to show that $\omega(A^n) \leq 1$.
He goes on by proving $\omega(A^n)\leq 1$ and stating that the theorem follows.
My problem is I can't see why $[\omega(A)]^n$ should be an upper bound for $\omega(A^n)$. That $[\omega(A)]^n \leq 1$ is trivial.
[1] "An elementary proof of the power inequality for the numerical radius" by Carl Pearcy in 1966
May everything stated above hold, then let us define
$$B = \frac{A}{\omega(A)}$$
for an $A \neq 0$.
If $C = 0$ then $\omega(C)=0$, hence $\omega(C^n)=[\omega(C)]^n=0$.
Now follows with $\omega(\lambda A) = |\lambda|\omega(A)$ that
$$\omega(B) = \omega\left(\frac{A}{\omega(A)}\right)=\frac{1}{\omega(A)}\cdot\omega(A) =1. $$
Hence $\omega(B^n) \leq 1$ and thus follows:
\begin{align}\omega(B^n)&=\omega\left(\frac{A^n}{[\omega(A)]^n}\right)\leq 1\\ &\Leftrightarrow \frac1{[\omega(A)]^n}\omega(A^n)\leq 1\\ &\Leftrightarrow \omega(A^n)\leq[\omega(A)]^n&\square\end{align}