summation - Alternating series of reciprocals of factorials

Consider the sum:
$$\frac{(-1)^2}{2!}+\frac{(-1)^3}{3!}+\frac{(-1)^4}{4!}+\ldots+ \frac{(-1)^n}{n!}.$$
Does there exist a nice closed form of such sum?

Answer

(UPDATED)
Concerning the finite sum :
$$\tag{1}S_n:=\sum_{k=2}^n\frac{(-1)^k}{k!}$$
we have simply : $$\tag{2} S_n=\frac 1{n!}\left[\frac{n!}e\right]=\frac{!n}{n!}$$
with $[x]$ the nearest integer (i.e. the round function) and $\,!n\,$ the number of derangements for $n$ elements
(added from Wood's answer (+1) in the related thread for additional properties).

To prove this directly you may use a method similar to the one proposed in this thread.

The idea is that the remaining terms of the Maclaurin expansion of $\,e^{-1}\;n!\,$ (after the $n$ first terms) are : $\;\displaystyle \frac {(-1)^{n+1}}{(n+1)}+ \frac {(-1)^{n+1}}{(n+1)(n+2)} +\cdots\;\ $ and thus with absolute value bounded by $\dfrac 12$ for $n>1$ that will disappear with the round function.