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.

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
$$\|a^n\|=\|\mathrm{id}_X^n\|_\infty=\|\mathrm{id}_X\|_\infty^n=\|a\|^n.$$
