x! - [(x-1)!+(x-2)!+(x-3)!...............2!+1!]
$\displaystyle 5! - \left( {\sum\limits_{k = 1}^4 {(5 - k)!} } \right) = 87$.
If $\displaystyle E_N=N! - \left( {\sum\limits_{k = 1}^{N-1} {(N - k)!} } \right)$ for $\displaystyle N\ge 2$ then we get $\displaystyle 1,~3,~15,~87,~567$ for the first five of those.
I have not seen a pattern yet.
See also the MathWorld article on the sum of factorials.
Also looking at this sequence site we see that this does not to yield much.
Ah, I figured out the mistake in my quickie guess-calculation.
a * b * c - b * c = (a-1) * b * c is true, but a * b * c - c = (a * (b-1) * c) is false. It should = ((a * b) - 1) * c, which doesn't simplify and gets worse the longer the expression is.