Nilpotent Element is Contained in Prime Ideals

Theorem
Let $R$ be a commutative ring with unity.

Let $\mathfrak p \subset R$ be a prime ideal of $R$.

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

Then $x \in \mathfrak p$.

Proof
Let $x$ be nilpotent in $R$ as asserted.

Then by definition:
 * $\exists n \in \Z_{>0}: x^n = 0$

But $0 \in \mathfrak p$ so:
 * $x^n \in \mathfrak p$

$x \notin p$.

Then by Complement of Prime Ideal of Ring is Multiplicatively Closed:
 * $x^n \notin \mathfrak p$

which contradicts the assertion that $x$ is nilpotent.

Thus:
 * $x \in \mathfrak p$