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How to prove that the number $1!+2!+3!+\dots+n!$ is never square?

Show that the sum $\sum_{k=1}^nk!\neq m^2$for any integer $m$, for $n\geq4$.

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    Using $m^2 \equiv 0,1,4 (\mathrm{mod}\;5)$.2012-12-07

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At $4$ the sum is $33$. Beyond $4$, every new term is divisible by $5$. So at $4$ or beyond, the sum is $\equiv 3\pmod{5}$. Nothing congruent to $3$ modulo $5$ can be a perfect square.