Prove that $\sum\limits_{k=1}^n (k!)^2$ is not a perfect square when $n\ge2$ 
Prove that $\displaystyle \forall n\geq 2, \sum_{k=1}^n (k!)^2$ is never a perfect square.

I'm far from well-read in number theory and I can't make any significant progress with this problem.
I tried to look at the sum $\text{mod}$ some small numbers, to no avail.

Hint: In base $10$, a square number can end only with digits $1, 4, 6, 9, 0$, or $5$. Now note that from $\left(5!\right)^{2}
 $ all factorials are multiple of $10$ and $$\sum_{k=1}^{4}\left(k!\right)^{2}=617.$$
