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.

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 $k_n=2^n$.
