Reduced Noetherian implies zero divisor in minimal prime

Statement

Suppose $A$ is a reduced Noetherian ring: in other words, it has no nilpotents, and every ideal is finitely generated. Then, any zero divisor in $A$ must be inside some minimal prime ideal of $A$.

(For non-Noetherianness, we can only say that every zero divisor must be inside some prime ideal, since minimal primes are not guaranteed to exist).

Proof

Suppose $a\in A$ is a zero divisor. Then, there exists $0\ne b\in A$ such that $ab=0$. Clearly, we have that $ab$ is in every minimal prime. If $a$ is not in any minimal prime, then $b$ must be in every minimal prime, hence $b$ must be in the intersection of all minimal primes. But the intersection of all minimal primes is the intersection of all primes, which is the nilradical. Since we're in a reduced ring, this is zero, so $b=0$, a contradiction.