<s>[INST] $1!+2!+3!+4!+5!+...+n!$ is a square of an integer 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?
 [/INST] 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
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