Show that the sum E(k=1 to n) k! not equals to m^2 for any integer m, for n>=4. E means sigma.

plzz, i need help. i am going to select best answer.......

Suppose to the contrary that
m^2 = 1! + 2! + 3! + 4! + ... + n! for some integer m.

Reducing mod 5 yields
m^2 = 1! + 2! + 3! + 4! + 0 (mod 5), since 5 | k! for k > 4
==> m^2 = 3 (mod 5).

This is a contradiction, because the squares mod 5 are 0, 1, 4.

I hope this helps!