The nilradical of an Artinian ring is a nilpotent ideal

Put $N=\mathrm{Nil}(R)$. The chain $N\supseteq N^2\supseteq N^3\supseteq \cdots$ is a descending chain of ideals, so it stabilizes: $N^n=N^{n+1}$ for some $n\ge 1$. If already $N^n=0$, then $N$ is nilpotent and there is nothing more to prove.

Assume instead that $N^n\mathrm{\ne }0$. Let $\mathcal{F}=\{\mathfrak{a}⊴R:\mathfrak{a}{N}^n\mathrm{\ne }0\}\mathrm{.}$ This family is nonempty because $(N^n)N^n=N^{2n}=N^n\mathrm{\ne }0$. By [L1], choose a minimal member $\mathfrak{a}$ of $\mathcal{F}$, and then choose $a\in \mathfrak{a}$ with $a{N}^n\mathrm{\ne }0$. Since $(a)\subseteq \mathfrak{a}$ and $(a)N^n\mathrm{\ne }0$, minimality gives $\mathfrak{a}=(a)$. Also $a{N}^n\subseteq \mathfrak{a}$, and because $N^n=N^{n+t}$ for every $t\ge 0$, one has $(a{N}^n)N^n=a{N}^{2n}=a{N}^n\mathrm{\ne }0$. So $a{N}^n\in \mathcal{F}$, whence minimality again gives $a{N}^n=\mathfrak{a}=(a)$. Therefore $a=ax$ for some $x\in N^n$.

The element $x$ lies in $N$, so The nilradical and reduced rings says that $x$ is nilpotent; say $x^m=0$. Iterating the identity $a=ax$ gives $a=a{x}^m=0$, contradicting $a{N}^n\mathrm{\ne }0$. Thus the assumption in step 2.1 is false, and the alternative left open in step 1.1 must hold: $N^n=0$.

Hence the nilradical of an Artinian ring is a nilpotent ideal.