How do you prove that a commutative C* algebra has no nontrivial nilpotents?

Question

In one of the responses to this question Commutative unital Banach algebra with nilpotent elements someone comments "But of course, you can't find a commutative C∗-algebra with nontrivial nilpotents."

How do you prove this?

Answer 1

In a commutative C$^*$-algebra, every element is normal. A normal nilpotent is zero, because a nilpotent has spectrum $\{0\}$ and the only normal with zero spectrum is zero.

Answer 2

Here is an alternative appealing directly to the C*-identity rather than the spectrum.

If $x$ is a normal element of a C*-algebra, i.e., $x^*x=xx^*$, then $$\|x^2\|^2=\|(x^2)^*x^2\|=\|(x^*x)^*(x^*x)\|=\|x^*x\|^2=(\|x\|^2)^2,$$

that is, $\|x^2\|=\|x\|^2$, which inductively implies $\|x^{2^k}\|=\|x\|^{2^k}$ for all $k\in\mathbb N$. If $x^n=0$ then some $x^{2^k}=0$ and thus $x=0$.