Is $1!^3+2!^3+3!^3+…n!^3$ a perfect square when $n>3$?

I noticed that $$1!^3=1$$ and $$1!^3+2!^3=9$$ are both perfect squares.

$$1!^3+2!^3+3!^3=225=15^2$$ is also a perfect square and even if $$n>5$$, it does end on $$9$$, so I think that $$1!^3+2!^3+3!^3+…n!^3$$ can be a perfect square.

Can $$1!^3+2!^3+3!^3+…n!^3$$ be a perfect square when $$n>3$$?

The first thing you should think of when you see such a problem is that $$n! \equiv 0 \pmod{k}$$ for all $$n \geq k$$. This means that $$S_n = \sum_{i = 1}^n (i!)^3$$ is eventually constant modulo $$p$$, and we just need to investigate whether $$S_n \pmod{p}$$ "ends" at a quadratic nonresidue modulo $$p$$. Through a small Python script, we get that $$p = 11$$ works for our purpose. That is,
\begin{align*} S_1 &= 1 \\ S_2 &= 9 \\ S_3 &= 225 \\ &\vdots\\ S_9 &= 47850402559694049 \equiv 7 &\pmod{11} \\ S_{10} &= 47832576242431694049 \equiv 6 &\pmod{11} \\ S_{11} &= 63649302669112063694049 \equiv 6 &\pmod{11} \\ S_{11 + k} &\equiv 6 &\pmod{11} \end{align*}
From here, simply note that $$6$$ is a quadratic nonresidue modulo $$11$$, so none of $$S_n$$ (for $$n \geq 11$$) is a square. You can check $$S_4$$ to $$S_{10}$$ manually (Euler style) that they are not squares.