$30 \cdot 26$ Let $w=7321 \cdot 7322 \cdot 7323 \cdot 7324+1$, then
(A) $w$ is a square number.
(B) $w$ is a prime number.
(C) $w$ is a cube number.
(D) $w$ is an even number.
(2nd China Tianjin "Chinese Youth Cup" Mathematics Invitational Competition, 1988)
[Solution] First consider the general case:
$$
\begin{aligned}
& n(n+1)(n+2)(n+3)+1 \\
= & \left(n^{2}+3 n\right)\left(n^{2}+3 n+2\right)+1 \\
= & \left(n^{2}+3 n\right)^{2}+2\left(n^{2}+3 n\right)+1 \\
= & \left(n^{2}+3 n+1\right)^{2},
\end{aligned}
$$

Thus, \( w=\left(7321^{2}+3 \cdot 7321+1\right)^{2} \) is a perfect square. Therefore, the answer is (A).