Can the Sum of Factorials Ever be a Complete Square?

[Karamata]

Homework Statement

Proof that for n\geq 5 expression \sum_{i=1}^{n}i! can't be a complete square

The Attempt at a Solution

Mathematical induction maybe?
n=5 \sum_{i=1}^{5}i!=1+2!+3!+4!+5!=152 OK
n\rightarrow n+1 \sum_{i=1}^{n+1}i!=\sum_{i=1}^{n}i!+(n+1)!
...
One guy said that

Complete square divided by 5 can give the remains 0,1 and 4, but not 3 like here.

But, I don't understand him.

Sorry for bad English.

 

[Oster]

1+2+6+24+120 = 153 right?
subsequent factorials are all divisible by 10, so dividing by 5 will always give the remainder 3.