Let $A$ be a Banach algebra with unit $e$ and let $a\in A$ be nilpotent. I want to show that the spectral radius $\rho(a)=0$.
We have that the spectral radius satisfies $$\rho(a)=\lim_{n\to\infty}\|a^{n}\|^{\frac{1}{n}}=\inf_{n\ge 1}\|a^{n}\|^{\frac{1}{n}}$$
Since $a$ is nilpotent, we have that $\exists n_{0}\in\mathbb{N}$ such that $a^{n_{0}}=0$. Since $n_{0}\ne 0$, it follows that $\rho(a)=0$.
I feel that this proof is incomplete though.
That's a good proof.
Here is another one. Assume that $a^m=0$. If you know that $$ \sigma(a)=\{f(a):\ f \text{ is a multiplicative functional }\}, $$ then for any such $f$ we have $f(a)^m=f(a^m)=f(0)=0$. So $f(a)=0$, and then $\sigma(a)=\{0\}$.
Yet another proof, without machinery: if $a^m=0$, then $1-a$ is invertible: indeed, the inverse is $1+a+a^2+\cdots+a^{m-1}$, because $$ (1-a)(1+a+a^2+\cdots+a^{m-1})=1. $$ For any $\lambda\ne0$ as $a/\lambda$ is nilpotent, $\lambda-a=\lambda(1-a/\lambda)$ is invertible. Thus $\lambda$ is in the resolvent of $a$. As this works for any $\lambda\ne0$, we get that $\sigma(a)=\{0\}$.