Is the sum of factorials of first $n$ natural numbers ever a perfect cube?

Question

If $S_n = 1! + 2! + 3! + \dots + n!$, is there any term in $S_n$ which is a perfect cube or out of $S_1$, $S_2$, $S_3$, $\dots S_n$ is there any term which is a perfect cube, where $n$ is any natural number.

Answer 1

All factorials above $8!$ have a factor of $27$ and $S_8 \equiv 9 \pmod {27}$ As there is no solution to $k^3 \equiv 9 \pmod {27}$, we cannot have $n \ge 9$. Then just checking $n$ up through $8$, only $S_1=1$ is a perfect cube.