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Consider numbers of the form
$$ u(n) = 1! + 2! + 3! +...+ n!. $$
b | b2 | Last digit of b2 |
0 | 0 | 0 |
1 | 1 | 1 |
2 | 4 | 4 |
3 | 9 | 9 |
4 | 16 | 6 |
5 | 25 | 5 |
6 | 36 | 6 |
7 | 49 | 9 |
8 | 64 | 4 |
9 | 81 | 1 |
This table shows that the squares of all integers between $0$ and $9$, and therefore all integers, end in either $1$, $4$, $5$, $6$ or $9$.
No perfect squares end in $3$. Therefore, $u(n)$ with $n > 4$ can never be a perfect square as it has been shown to always end in $3$.
All that remains is to find $u(n)$ for all integers of $4$ or less:
Only $u(1)$ and $u(3)$ are perfect squares, and so they are the only sums of factorials to be perfect squares.