Nilpotent Element is Zero Divisor

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

Suppose further that $R$ is not the null ring.

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

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

Proof
Let $0_R$ be the zero of $R$.

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

If $n = 1$, then $x = 0_R$.

By hypothesis, $R$ is not the null ring, so we may choose $y \in R \mathop \backslash \left\{ {0} \right\}$.

Now by Ring Product with Zero we have
 * $y \circ x = y \circ 0_R = 0_R$

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

If $n \ge 2$, 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 $R$.

Remark
Note that in the case when $R$ is the null ring the result is false.

This is because $0_R$ is nilpotent element in the null ring, but is not a zero divisor.

This is in contrast with non-null rings $R$: in this case $0_R$ is both nilpotent and a zero divisor.