Sum of factorials and perfect squares

Question

Is there a formula that explicits which $n$ makes the following a perfect square?

$ \sum_{i=1}^{n}i! = a^{2}, \quad a \in \mathbb{Z}$

Answer 1

Guide:

$$0^2 \equiv 0\pmod{10}$$ $$(\pm 1)^2 \equiv 1\pmod{10}$$ $$(\pm 2)^2 \equiv 4\pmod{10}$$ $$(\pm 3)^2 \equiv 9\pmod{10}$$ $$(\pm 4)^2 \equiv 6\pmod{10}$$

Find the unit digit of $$ \sum_{i=1}^{n}i! $$

for $n\ge4$.

Hint:

$5!$ is a multiple of $10$.

Solution for all possible $n$:

$n$ is either $1$ or $3$.