Prove $1! + 2! + 3! + \ldots + n! =y^3$ has only one solution in the set of natural numbers? I actually know that the above equation is true for $n=1$ and $y=1$ but am unable to prove it for the entire set of natural numbers. Can anyone please help me solve this in a simple way?

Hint. Let
$$a_n = \sum_{k=1}^{n} k!.$$
Then $a_n$ is divisible by 3 for all $n \geq 2$, and we have $a_n \equiv 0 \pmod{27}$ only when $n=7$.
