TITLE: How to prove that the number $1!+2!+3!+...+n!$ is never square?
QUESTION [32 upvotes]: How to prove that the number $1!+2!+3!+...+n! \ \forall n \geq 4$ is never square?
I was told to count permutations but I cannot figure out what we are permuting.... Thanks!

REPLY [4 votes]: You can also use mod $7$ to solve this.
Firstly, manually checking $n=1,2,3,4,5$ gives only the solutions $n=1$, $n=3$.
If $n\ge 6$, then $$1!+2!+\cdots+n!$$
$$\equiv 1!+2!+3!+4!+5!+6!\pmod{7}$$
$$\equiv 5\equiv m^2\pmod{7},$$
contradiction, because $5$ is not a quadratic residue mod $7$ (notice $(7k\pm 1)^2\equiv 1\pmod{7}$, $(7k\pm 2)^2\equiv 4\pmod{7}$, $(7k\pm 3)^2\equiv 2\pmod{7}$).
mod $13$ would also work analogously. If $n\ge 12$, then $$1!+2!+\cdots+n!$$
$$\equiv 1!+2!+\cdots+12!\equiv 11\pmod{13}$$
and $11$ is not a quadratic residue mod $13$.
Using this method mod $9$ or mod $11$ wouldn't work (the residues would be $0$, which is a square mod every integer).
mod $17$ would also work -- the residue would be $12$, which is not quadratic mod $17$.