Can $!1+!2+!3+\cdots+!n$ be a perfect power if $n\geq3$?
Note that $!n$ is a subfactorial.
I do know that $1!+2!+3+\cdots+n!$ is only a perfect power if $n=1, 3$, since when $n\geq9, 1!+2!+3!+\cdots+9!=9\pmod{27}$, so it cannot be a perfect cube, or any higher perfect prime power, and this is never a perfect square if $n\geq5$.
But I don’t see any pattern to $!1+!2+!3+\cdots+!n$, so I cannot know if this can be a perfect power if $n\geq3$.
For example, $!1+!2+!3+\cdots+!16\equiv1\pmod{5,7},0\pmod{9},0\pmod{16}$, so it can a perfect square or a higher perfect power.
On the other hand, $!1+!2+!3+\cdots+!17\equiv8\pmod{9}$, so it is not a perfect square, but it can be a odd perfect power.
Are there any ways to find the remainder of $!1+!2+!3+\cdots+!n$ when divided by $p\geq n$ to determine if $!1+!2+!3+\cdots+!n$ can be a perfect power?