$1!+2!+\ldots+n!$ cannot be the square of a positive integer

I have to prove that $1!+2!+\ldots+n!$ cannot be the square of a positive integer, $\forall n\geq4$.

I've tried to do this with induction, but I don't seem to reach any satisfactory conclusion.

Any hint will do.

Solution 1:

Hint: First, see if you can show that no perfect square can be equivalent to $3 \pmod{10}$.

From here, note that $n! \equiv 0 \pmod{10}$ for all $n \geq 5$.

Hence, we get:

$$\sum\limits_{k=1}^nk! \equiv 1! + 2! + 3! + 4! \equiv 33 \equiv 3 \pmod{10}$$

Solution 2:

See that 1 ! + 2 ! + 3 ! + 4 !=33. And from 5! onwards all the other numbers will end with 0 as its last digit. So when you take the sum upto n! with n greater than equal to 4 you can see that the last digit of the sum will be 3. So there is no postive integer whose square ends up with 3 as the last digit or unit place's digit. So the sum cannot be a square of a positive integer.