Is $7$ the largest possible value of $n$ for which $(1!+2!+3!+…n!)+16$ is a perfect power?

Question

Is $7$ the largest possible value of $n$ for which $(1!+2!+3!+…n!)+16$ is a perfect power?

I noticed that $(7!+6!+5!+4!+3!+2!+1!)+16=77^2$ is a perfect power, and I don’t know if that is the largest possible value of $n$ for which $(1!+2!+3!+…n!)+16$ is a perfect power?

I can conclude that $(1!+2!+3!+…n!)+16$ is never a perfect cube if $n\geq6$, since it is equal to $4\pmod{9}$, but I don’t know if $(1!+2!+3!+…n!)+16$ can be a perfect square.

Since the last digit of $(1!+2!+3!+…n!)+16$ is $9$, I think it can be a perfect square if $n\geq8$, since the sum is $1\pmod{8}$, and $1\pmod{3}$.

I can also conclude that $(1!+2!+3!+…n!)+16$ cannot be a perfect 5th power if $n\geq10$, since $29$ is not a 5th power residue $\pmod{100}$.

Other than $3, 4, 5, 7$, are there any values of $n$ such that $(1!+2!+3!+…n!)+16$ is a perfect power?

Edit:

• Since $3$ is not a 7th power residue $\pmod{29}$, $(1!+2!+3!+…n!)+16$ cannot be a perfect 7th power if $n\geq28$.

Answer 1

$$(1! + 2! + 3! + \ldots + n!) + 16 \equiv 14 \pmod{49}$$ for $n \ge 13$. So $(1! + 2! + 3! + \ldots + n!) + 16$ is not a perfect power if $n \ge 13$.

Answer 2

A proof for perfect even power $m^{2k}$

Assume, on the contrary, that $n>7$, then $n! \equiv 0 \quad \pmod {32}$ and

$$ \begin{aligned} & 1 !+2 !+3 !+ 4!+5!+6! +7 !+8 !+\cdots+n !+16 \\ \equiv & 1+2+6+24+120+720+5030+16 \\ \equiv & 5919 \\ \equiv & -1(\bmod 32) \end{aligned} $$

which contradicts to the fact that $m^{2k}\not \equiv -1\quad \pmod {32}$ for any integers $m$ and $k$.

Therefore $1 !+2 !+3 !+4!+5!+6!+7 !+8 !+\cdots+n !+16$ is never a perfect even power for $n>8$. On the other hand, $1 !+2 !+3 !+\ldots+7 !+16=77^2$ reveals that $7$ is largest integer to make the sum a perfect even power.

Unfortunately, this method can’t be applied to perfect odd powers.