Solution:

Given any prime $p$ and positive integer $x$, let $v_{p}(x)$ denote the highest power of $p$ dividing $x$. We claim that Jane cannot write 2024 such numbers as that would imply that $1! \cdot 2! \cdots 2024!$ is the square of the product of the 2024 numbers. Let $p$ be a prime and $k$ be a natural number such that $k < p$, $k p \leq 2024$, and $(k+1) p > 2024$. Then note that
$$v_{p}(1! \cdot 2! \cdots 2024!) = (2024 - p + 1) + (2024 - 2p + 1) + \ldots + (2024 - k p + 1)$$
In particular, let $p$ be in $\left(\frac{2024}{4}, \frac{2024}{2}\right)$. By Bertrand's Postulate, such a prime $p$ exists (and $p$ must also be odd). Further, the corresponding $k$ is either 2 or 3. Either way, $v_{p}(1! \cdot 2! \cdots 2024!)$ is odd from the above formula, and so $1! \cdot 2! \cdots 2024!$ cannot be a perfect square.

Solution:

As in the first solution, we prove $1! \cdot 2! \cdots 2024!$ is not a perfect square. To do this, note that we can rewrite the product as $(1!)^{2} \cdot 2 \cdot (3!)^{2} \cdot 4 \cdots (2023!)^{2} \cdot 2024$ which is
$$2 \cdot 4 \cdots 2024 \cdot (1! \cdot 3! \cdots 2023!)^{2} = 1012! \cdot \left(2^{1012} \cdot 1! \cdot 3! \cdots 2023!\right)^{2}$$
so it is sufficient to verify $1012!$ is not a perfect square. This can be verified by either noticing the prime $1009$ only appears as a factor of $1012!$ once, or by evaluating $v_{2}(1012!) = 1005$.