Does there exist a positive integer value of $n$ such that the sum of $1!^n+2!^n+3!^n+\cdots+2024!^n$ is a perfect power?
I know that $n\neq1$, as the last digit of the sum when $n\geq4$ is $3$ and no perfect square ends with $3$, and it cannot be a perfect power since $3^3$ does not divide the sum when $n\geq8$.
I don’t know about when $n=2$, since the only sure point when $1!^2+2!^2+3!^2+\cdots+n!^2$ cannot be a perfect power is when $n\geq1248828$,but I don’t have any idea if $1!^2+2!^2+3!^2+\cdots+2024!^2$ is a perfect power.
If $n=3$, then the sum isn’t a perfect square when $n\geq10$ as it is equal to $6\pmod{11}$, and it cannot be a perfect power since $3^3\nmid1!^3+2!^3+3!^3+\cdots+n!^3$ when $n\geq5$.
If $n=5,7$, then it is not a perfect power since $3^2$ dosen’t divide the sum.
Also, for all $n$ that is even, the sum can’t be a perfect square because the sum is equal to $2\pmod{3}$.
Note also that if $n$ is odd, then $1!^n+2!^n+3!^n+\cdots+2024!^n$ isn’t a perfect powers because there is some power of three that does not divide the sum after some point(in some cases, $9$ even dosen’t divide the sum), and forces the sum to be perfect square in order to be a perfect power. But also, the sum will “land” on a prime, where $1!^n+2!^n+3!^n+\cdots+2024!^n\equiv a\pmod{p}$, and $a$ isn’t a quadratic residue $\pmod{p}$.
So I suspect that $1!^n+2!^n+3!^n+\cdots+2024!^n$ is never a perfect square unless $n$ is divisible by $3$.
Now, it forces us that $n$ has to be even in order for $1!^n+2!^n+3!^n+\cdots+2024!^n$ to be a higher odd prime perfect power.
I used Pari GP and checked the all values of $n\leq10^{4}$ to see if there is a value of $n$ such that $1!^n+2!^n+3!^n+\cdots+2024!^n$ is a perfect power, but so far, none has been found.
Edit: This answer already ruled out about the sum being a perfect square.
Can $1!^n+2!^n+3!^n+\cdots+2024!^n$ be a perfect power?
Best Answer
As you already demonstarted, the sum can only be a square $(\bmod 3)$ if $n$ = $1(\bmod 2) \implies n=1,3,5,7,9,11(\bmod 12)$
To be a square $(\bmod 5) \implies n = 3(\bmod 4) \implies n= \not1,3,\not5,7,\not9,11(\bmod 12)$
To be a square $(\bmod 9) \implies n=1,3(\bmod 6)\implies n=\not1,3,\not5,7,\not9,\not11(\bmod 12)$
To be a square $(\bmod 13) \implies n=1,5,11 (\bmod 12) \implies n = \not1,\not3,\not5,\not7,\not9,\not11 (\bmod 12)$ $$\implies \sum_N^{N\ge12} N!^n\not = k^2 $$