8. If one element is deleted from the set $\{1!, 2!, \cdots, 100!\}$, and the product of the remaining elements is exactly a perfect square, then the deleted element is $\qquad$ .
8. 50 !.

Let $A=1!\times 2!\times \cdots \times 100!$.
Since $(2 k)!=(2 k) \cdot(2 k-1)!$, we have
$$
\begin{aligned}
A & =(1!)^{2} \times 2 \times(3!)^{2} \times 4 \times \cdots \times(99!)^{2} \times 100 \\
& =(1!\times 3!\times \cdots \times 99!)^{2}(2 \times 4 \times 6 \times \cdots \times 100) \\
& =(1!\times 3!\times \cdots \times 99!)^{2}\left(2^{25}\right)^{2} \times 50!.
\end{aligned}
$$

Thus, the product of the remaining elements after removing 50! from $A$, $\frac{A}{50!}$, is a perfect square.
Next, we prove that removing 50! is unique.
If we can also remove $k!$ such that $\frac{A}{k!}$ is a perfect square, then when $50<k<100$, $51 \times 52 \times \cdots \times k$ cannot be a perfect square (the exponent of 53 is 1), and $51, 51 \times 52$ are not perfect squares.

Since 47 is a prime number, when $1 \leqslant k \leqslant 46$, $(k+1) \times \cdots \times 49 \times 50$ cannot be a perfect square, and $48 \times 49 \times 50, 49 \times 50, 50$ are not perfect squares.
In summary, the element 50! to be removed is unique.