Nilpotent Element is Zero Divisor

Theorem
Let $\left({R, +, \circ}\right)$ be a ring.

Let $x \in R$ be a nilpotent element of $R$.

Then $x$ is a zero divisor in $R$.

Proof
Let $0_R$ and $1_R$ be the identities for $+$ and $\circ$ in $R$ respectively.

By hypothesis, there exists $n \geq 1$ such that $x^n = 0_R$.

If $n = 1$, then we have:
 * $\displaystyle 1_R \circ x = x = 0_R$

so $x$ is a zero divisor in $R$.

If $n \geq 2$, define $y = x^{n-1}$. Then we have:
 * $\displaystyle y \circ x = x^{n-1} \circ x = x^n = 0_R$

so $x$ is a zero divisor in $R$.