12. After removing one element from the set $\{1!, 2!, \cdots, 24!\}$, the product of the remaining elements is exactly a perfect square.
12.12!.

From $(2 k)!=(2 k)(2 k-1)!$, the product can be transformed into
$$
\begin{array}{l}
24 \times(23!)^{2} \times 22 \times(21!)^{2} \times \cdots \times 4 \times(3!)^{2} \times 2 \\
=\left[2^{6} \times(23!) \times(21!) \times \cdots \times(3!)\right]^{2} \times(12!) .
\end{array}
$$

Therefore, the factor 12 ! is extracted.