We can see, from 7! onwards the terms are divisible by 14.

∵7!=7×6×5×4×3×2×1

=(7×2)×6×5×4×3

=14×360

So, we should consider up to 6! term to obtain the remainder.

Now,

1! + 2! + 3! + 4! + 5! + 6!

= 1 + 2 + 6 + 24 + 120 + 720

= 873

On dividing 873 by 14, we get 5 as remainder.

∵14×62=868

and 873-868=5