# Can $!1+!2+!3+\cdots+!n$ be a perfect power?

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?

• Since $!1+!2=1$, maybe change the hypothesis to $n\ge3$. Sep 27, 2023 at 9:59
• Usually, $1$ is not considered to be a perfect power , but we can begin with $2$ since $!1=0$. No perfect power upto $n=1\ 000$ (brute force with PARI/GP). Sep 29, 2023 at 22:55
• What is the def of $!n$? Oct 6, 2023 at 15:37
• @C.F.G: "Note that $!n$ is a subfactorial." Oct 6, 2023 at 16:04