Nilradical of a ring

Definition: Let $A$ be a commutative ring. We define the nilradical of $A$ to be
$$Nil(A) := \{ x \in A ~|~ x^n = 0 \text{ for some } n \geq 1 \}$$

Then we get a nice equivalence

$$Nil(A) = \bigcap_{\mathfrak{p} \in Spec(A)} \mathfrak{p}$$

Let $x \in A$ be nilpotent, i.e. $x^n = 0$ for some $n \in \mathbb{N}$ then $x^n = 0 \in \mathfrak{p}$ for any prime ideal $\mathfrak{p}$. By induction, this shows that $x \in \mathfrak{p}$.

Now suppose that $x \notin A$, then $x^n \neq 0$ for any $n \in \mathbb{N}$. In particular, the following collection of ideals is non-empty,
$$S = \{ \mathfrak{A} ~|~ \mathfrak{A} \text{ ideal of } A \text{ and } x^n \notin \mathfrak{A} \text{ for all } n \}$$

because the zero ideal $(0)$ is in $S$. If we have a chain of ideals
$$\mathfrak{A}_1 \subset \mathfrak{A}_2 \subset \cdots$$
then we can take the union of all terms in the chain, call it, $\mathfrak{B}$. Then, it turns out that $\mathfrak{B}$ is an ideal. Hence we can use the Zorn's lemma to find the maximal element, say, $\mathfrak{p}$. We claim that $\mathfrak{p}$ is a prime ideal.

Suppose for contradiction that there exists $a,b \notin \mathfrak{p}$ such that $ab \in \mathfrak{p}$. We have a proper inclusion
$$\mathfrak{p} \subset \mathfrak{p} + (a) \text{ and } \mathfrak{p} \subset \mathfrak{p} + (b)$$
by maximality, there exists $n, m$ such that $x^n \in  \mathfrak{p} + (a)$ and  $x^m \in  \mathfrak{p} + (b)$. As a consequence,
$$x^{n+m} \in (\mathfrak{p} + (a)) ( \mathfrak{p} + (b)) \subset \mathfrak{p}$$
which is a contradiction. This shows that $\mathfrak{p}$ is a prime ideal. As $\mathfrak{p}$ is an element in $S$, it follows that $x \notin \mathfrak{p}$, hence
$$x \notin \bigcap_{\mathfrak{p} \in Spec(A)} \mathfrak{p}$$
This completes the proof.