Find the values of $n\geq 1$ for which $1!+2!+\cdots+n!$ is a square.
I wanted to go through some small $n$..
$1!=1$
$1!+2!=3$ not a square
$1!+2!+3!=9$ a square
$1!+2!+3!+4!=33$ not a square
$1!+2!+3!+4!+5!=153$ not a square
$1!+2!+3!+4!+5!+6!=873$ not a square
Here comes a pattern... All $n!$ for $n\geq 5$ ends with $0$ because $5\cdot 2$ is factor of $5!$.
So, $1!+2!+\cdots+n!$ for $n\geq 5$ ends with $3$ so can not be a square..
See that unit digit of a square can only be one of $0,1,4,5,6,9$..
So, only possible $n$ are $1,3$
I just want to know if there are any gaps.