= Solution
For a <self-adjoint C-star element> $h$, the <C-star identity> gives $\|h^2\|=\|h\|^2$. If $x$ is a <normal C-star element>, then
$$
\|x^2\|^2=\|(x^*)^2x^2\|=\|(x^*x)^2\|=\|x^*x\|^2=\|x\|^4.
$$
All powers are <normal C-star elements>, so induction gives $\|x^{2^j}\|=\|x\|^{2^j}$. The general <spectral radius formula> now gives $r(x)=\|x\|$.
One can also see the needed growth estimate directly. For $R>r(x)$ the <resolvent Cauchy coefficient formula> gives
$$
x^m=\frac1{2\pi i}\int_{|z|=R}z^m(z1-x)^{-1}\,dz,\qquad\|x^m\|\le C_RR^{m+1}.
$$
The formula follows by the <Neumann series> on a larger circle and <contour deformation> through the resolvent annulus. Put $m=2^j$, take $m$th roots and let $j\to\infty$ to get $\|x\|\le R$. Let $R\downarrow r(x)$; the reverse inequality already follows from the <norm> bound on the <algebra spectrum>. Thus
$$
\boxed{r(x)=\|x\|}.
$$
This proves <spectral radius norm equality for normal elements> from the <norm> identity.
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