For a hermitian element $a$ in a $C^*$-algebra, show that $\|a^{2n}\| = \|a\|^{2n}$ Let $\mathcal{A}$ be a $C^*$-algebra. Suppose that $a \in \mathcal{A}$ with the property that $a^* = a$ (that is, suppose that $a$ is hermitian). 

I would like to show that $\|a^{2n}\| = \|a\|^{2n}$ for all $n \ge 1$.

This fact is stated in Conway's A Course in Functional Analysis, and I'm having trouble proving it. Here's what I have so far: the $n =1 $ base cases just uses the basic $C^*$-algebra property:
$$\|a^2\| = \|a^* a \| = \|a\|^2.$$
But now I get stuck on the induction step (assume we have $\|a^{2n}\| = \|a\|^{2n}$, show that $\|a^{2n + 2}\| = \|a\|^{2n + 2}$). I have been trying to give myself some insight by studying small cases, for instance, $n = 3$:
$$\|a^6\| = \|(a^3)^2\| = \|a^3\|^2,$$
where the last equality follows from the base case. But then I end up with an odd exponent inside the norm. This is what is really giving me trouble.
Hints or solutions are greatly appreciated.
 A: This is true for all powers, even or odd.  
Proposition 1.11e on page 234 says that $\|a\|=r(a)$ for any Hermitian $a$, where $r(a) $ is the spectral radius of $a$, that is $r(a) = \lim_{k\to\infty} \|a^k\|^{1/k}$. 
Since $a^{n}$ is also Hermitian, we have 
$$
\|a^{n}\| =r(a^n) = \lim_{k\to\infty} \|a^{kn}\|^{1/k} = \left(\lim_{k\to\infty} \|a^{kn}\|^{1/(kn)}\right)^n = r(a)^n = \|a\|^n \tag{1}
$$

For completeness, the proof of $\|a\|=r(a)$ is based on the equality $\| a^{2^n}\|=\|a\| ^{2^n}$ which, as Kaladin noted in a comment, is shown by induction: 
$$\| a^{2^n}\| = \left\| \left(a^{2^{n-1}}\right)^2\right\|  = \| a^{2^{n-1}}\| ^2 = \dots =\|a\|^{2^n} \tag{2}$$
Since the limit $r(a) = \lim_{k\to\infty} \|a^k\|^{1/k}$ is known to exist (page 197), one can use (2) to calculate it along the subsequence $k_n=2^n$.
A: Alternatively, let $A$ be the $C^*$-algebra generated by $a$. Since $a$ is hermitian (or more generally normal), $A$ is commutative and isometrically isomorphic under the Gelfand transform to $(C(X),\|\cdot\|_\infty)$, where $X=\sigma(a)$ is the spectrum of $a$, and $a$ corresponds to the identity function of $X$; thus
$$\|a^n\|=\|\mathrm{id}_X^n\|_\infty=\|\mathrm{id}_X\|_\infty^n=\|a\|^n.$$
