Here is one way method which requires no knowledge of Noetherian rings.
(1) Reduce to the case that $A$ is reduced, since $A$ Noetherian implies $A/\sqrt{0}$ Noetherian and the minimal primes of $A$ and $A/\sqrt{0}$ are in bijection.
(2) For each minimal prime $P_\alpha$, use that $A_{P_\alpha}$ is a field and that $P_\alpha$ is finitely generated (by the Noetherian hypothesis) to produce an $s_\alpha \in A \setminus P_\alpha$ such that $s_\alpha P_\alpha = 0$.
(3) Let $I$ be the ideal generated by the $s_\alpha$. By construction, $I$ is not contained in any minimal prime ideal. Check that this implies that $Ann(I) = 0$, i.e. if $aI = 0$ then $a = 0$.
(4) By the Noetherian hypothesis, $I$ is finitely generated, say by $s_1, \ldots, s_n$. Let $P_i$ be the corresponding minimal prime ideals. Check that $I \prod_{i=1}^n P_i = 0$, and therefore by the previous observation, $\prod_{i=1}^n P_i = 0$.
(5) Check that whenever $A$ is a ring in which some finite product of prime ideals is equal to $0$, $A$ necessarily has only finitely many primes.