proof - Show that $1! +2! +3!+\cdots+n!$ is a perfect power if and only if $n=3$

$1!+2!+\dots + 8! \equiv 9\pmod{27}$, and any additional term you add is $\equiv 0 \pmod{27}$.

So there can be no larger power than $2$ when $n \ge 8$, as $3$ is a factor but $27$ is not.

There can be no square either for $n> 3$, as $1!+2!+3!+4!=33$ and all additional terms have last digit $0$, and no square ends in a $3$.

The remaining finite cases are easily checked.

Note that,$$1! + 2! + 3! + 4! = 33$$ $$1! + 2! + 3! + 4! + 5! = 153$$ $$1! + 2! + 3! + 4! + 5! + 6! = 873 \ldots $$ The last digit of the numbers is $3$ (This is happening because for $n>4$ the last digit of $n!$ is $0$). Now for a number to be a perfect square the last digit should be one of the digits $1,4,5,6,9$. Hence for $n>3$, $\sum\nolimits_{i = 1}^n {i!} $ cannot be a perfect square.