TITLE: Norm of self-adjoint member of $C^*$-algebra
QUESTION [5 upvotes]: This question arose from the proof of proposition $1.11(e)$ in chapter $8$ of  John B. Conway's A Course in Functional Analysis.  This portion of the proposition can be stated:

Let $\mathscr{A}$ be a $C^*$-algebra, and let $a\in\mathscr{A}$ be given. If $a=a^*$, then $\|a\|=r(a)$.

(Here, $r(a)$ denotes the spectral radius of $a$.)
The proof, as stated in the book, proceeds as follows:

Since $a^*=a$, $\|a^2\|=\|a^*a\|=\|a\|^2$; by induction, $\|a^{2n}\|=\|a\|^{2n}$ for $n\geq1$  That is, $\|a^{2n}\|^{1/2n}=\|a\|$ for $n\geq1$.  Hence $r(a)=\lim\|a^{2n}\|^{1/2n}=\|a\|$.

Now I was able to show by induction that
$$ \|a^{2^n}\|=\|a\|^{2^n} \qquad (n\geq1),$$
from which the result follows, but I could not prove it as it is stated in the book.  
So my question is:  How can we prove (presumably by induction) that $\|a^{2n}\|^{1/2n}=\|a\|$ for $n\geq1$? Is this simply an error in the book, or can it be done?

REPLY [3 votes]: This was corrected in a later edition ...
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