 Solution

First the C-star identity and submultiplicativity give $\Vert a{\Vert }^2=\Vert a^{\ast }a\Vert \le \Vert a^{\ast }\Vert \Vert a\Vert$, hence $\Vert a\Vert \le \Vert a^{\ast }\Vert$ when $a\ne 0$. Applying this to $a^{\ast }$ gives equality, so the involution is isometric and continuous. Also $\Vert 1\Vert =\Vert 1{\Vert }^2$ and $1\ne 0$, yielding $\Vert 1\Vert =1$.

For a Normal element of a C-star algebra $x$, put $b=x^{\ast }x$, which is a self-adjoint C-star element. Normality gives $(x^{\ast }{)}^2{x}^2=b^2$, and the C-star identity applied to $x^2$ and to $b$ givesThus $\Vert x^2\Vert =\Vert x{\Vert }^2$. Every power of $x$ is normal because $x$ commutes with $x^{\ast }$, so iteration yieldsThe spectral radius formula has a limit along all positive integers, and along this subsequence its root norms are exactly $\Vert x\Vert$. ThereforeThis proves spectral radius norm equality for normal elements directly from the defining norm identity, rather than assuming a spectral representation in advance.