= Solution
First the <C-star identity> and <submultiplicativity> give $\|a\|^2=\|a^*a\|\le\|a^*\|\|a\|$, hence $\|a\|\le\|a^*\|$ when $a\ne0$. Applying this to $a^*$ gives equality, so the <involution> is isometric and continuous. Also $\|1\|=\|1\|^2$ and $1\ne0$, yielding $\|1\|=1$.
For a <Normal element of a C-star algebra> $x$, put $b=x^*x$, which is a <self-adjoint C-star element>. Normality gives $(x^*)^2x^2=b^2$, and the <C-star identity> applied to $x^2$ and to $b$ gives
$$
\|x^2\|^2=\|(x^*)^2x^2\|=\|b^2\|=\|b\|^2=\|x\|^4.
$$
Thus $\|x^2\|=\|x\|^2$. Every power of $x$ is normal because $x$ commutes with $x^*$, so iteration yields
$$
\|x^{2^j}\|=\|x\|^{2^j}\qquad(j\ge0).
$$
The <spectral radius formula> has a limit along all positive integers, and along this subsequence its root <norms> are exactly $\|x\|$. Therefore
$$
\boxed{r(x)=\|x\|.}
$$
This proves <spectral radius norm equality for normal elements> directly from the defining <norm> identity, rather than assuming a spectral representation in advance.
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