5. By definition, $n!=1 \cdot 2 \cdot 3 \cdot \ldots \cdot n$. Which factor should be removed from the product $1! \cdot 2! \cdot 3! \cdot \ldots \cdot 20!$ so that the remaining product becomes a square of some natural number?
Solution. Note that

$$
\begin{aligned}
& 1!\cdot 2!\cdot 3!\cdot 4!\cdot \ldots \cdot 20!=(1!\cdot 2!) \cdot(3!\cdot 4!) \cdot \ldots \cdot(19!\cdot 20!)= \\
= & (1!\cdot 1!\cdot 2) \cdot(3!\cdot 3!\cdot 4) \cdot(5!\cdot 5!\cdot 6) \cdot \ldots \cdot(17!\cdot 17!\cdot 18) \cdot(19!\cdot 19!\cdot 20)= \\
= & (1!)^{2} \cdot(3!)^{2} \cdot(5!)^{2} \cdot \ldots \cdot(19!)^{2} \cdot(2 \cdot 4 \cdot 6 \cdot 8 \cdot \ldots \cdot 18 \cdot 20)= \\
= & (1!)^{2} \cdot(3!)^{2} \cdot(5!)^{2} \cdot \ldots \cdot(19!)^{2} \cdot(2 \cdot(2 \cdot 2) \cdot(3 \cdot 2) \cdot \ldots \cdot(10 \cdot 2))= \\
= & (1!\cdot 3!\cdot \ldots \cdot 19!)^{2} \cdot 2^{10} \cdot(1 \cdot 2 \cdot 3 \cdot \ldots \cdot 10)=(1!\cdot 3!\cdot \ldots \cdot 19!)^{2}\left(2^{5}\right)^{2} \cdot 10!
\end{aligned}
$$

It is obvious that the first two factors are squares, so if we remove 10!, a square will remain. It is easy to see that removing other factors does not yield the desired result.

Answer. 10!