$1!+2!+3!+4!+5!+...+n!$ is a square of an integer

Question

At what n is the number $1!+2!+3!+4!+...+n!$ a square of another integer number? All in all I have found $2$ values: $1$ and $3$. I think they are the only ones, but the only thing left is to prove it. How?

Answer 1

Let $1!+2!+3!+\cdots \cdots +n! = y^2$

$\star$ if $n=1\;,$ then $1=y^2$ (True)

$\star$ if $n=2\;,$ then $1+2=3=y^2$ (False)

$\star$ if $n=3,$ then $1+2+6=9=y^2$ (True)

$\star$ if $n=4,$ then $1+2+6+24=33=y^2$(False)

for $n\geq 5,$ then L .H .S end with $3$ and we now that square of any integer does not have last digit $3$

so we have only $n=1,n=3$ for which $y^2$ is a perfect square quantity