3. After deleting one element $\qquad$ from the set $\{1!, 2!, \cdots, 24!\}$, the product of the remaining elements is exactly a perfect square.
The product of the elements in the set is
$$
\begin{array}{l}
\prod_{i=1}^{12}[(2 i)!\times(2 i-1)!] \\
=\left[\prod_{i=1}^{12}(2 i-1)!\right]^{2} \times 2^{12} \times 12!.
\end{array}
$$

Therefore, the element to be deleted is 12!.