[proofplan] We use the definition of nilpotence to choose a positive exponent for which a power of $a$ is zero, then choose the least such exponent. If that exponent is $1$, the conclusion is immediate. If it is larger than $1$, the equation $a^n = a a^{n-1} = 0_R$ and the zero-product property of an [integral domain](/page/Integral%20Domain) force either $a = 0_R$ or a smaller positive power of $a$ is zero; the latter contradicts minimality. [/proofplan] [step:Choose the least positive exponent that kills $a$] Assume $a \in R$ is nilpotent. Define the set of nilpotence exponents of $a$ by \begin{align*} E := \{m \in \mathbb{N} : a^m = 0_R\}. \end{align*} Since $a$ is nilpotent, $E$ is nonempty. By the well-ordering property of $\mathbb{N}$, there exists $n \in E$ such that $n \le m$ for every $m \in E$. Thus $a^n = 0_R$, and no positive integer smaller than $n$ has this property. [/step] [step:Use the zero-product property to rule out a minimal exponent greater than one] If $n = 1$, then $a = a^1 = 0_R$, as required. Suppose instead that $n > 1$. Define $b := a^{n-1} \in R$. Since $n > 1$ and $n \in \mathbb{N}$, the integer $n-1$ belongs to $\mathbb{N}$. We have \begin{align*} a b = a a^{n-1} = a^n = 0_R. \end{align*} Because $R$ is an integral domain, it has no zero divisors. Hence $a = 0_R$ or $b = 0_R$. If $b = 0_R$, then $a^{n-1} = 0_R$, so $n-1 \in E$, contradicting the minimality of $n$. Therefore $a = 0_R$. [guided] The purpose of choosing $n$ minimally is to make the alternative $a^{n-1} = 0_R$ impossible. There are two cases. First, if $n = 1$, then the equation $a^n = 0_R$ says exactly $a^1 = 0_R$, so $a = 0_R$. Now suppose $n > 1$. Since $\mathbb{N}$ starts at $1$, the inequality $n > 1$ implies $n-1 \in \mathbb{N}$. Define $b := a^{n-1} \in R$. Using the recursive law for powers in a ring, we factor the zero power as \begin{align*} a b = a a^{n-1} = a^n = 0_R. \end{align*} The defining zero-product property of an integral domain says that whenever $x,y \in R$ satisfy $xy = 0_R$, then $x = 0_R$ or $y = 0_R$. Applying this to $x := a$ and $y := b$, we obtain $a = 0_R$ or $b = 0_R$. If $a = 0_R$, we are done. If $b = 0_R$, then $a^{n-1} = 0_R$, so $n-1$ is an element of the set \begin{align*} E = \{m \in \mathbb{N} : a^m = 0_R\}. \end{align*} But $n-1 < n$, contradicting the choice of $n$ as the least element of $E$. Thus the second alternative cannot occur, and the only possible conclusion is $a = 0_R$. [/guided] [/step]