Now, how do we get a formula for these numbers out of this? First note that
$$exp((u-1)z) = 1 + (u-1)z + {(u-1)^2 \over 2!} z^2 + {(u-1)^3 \over 3!} z^3 + \cdots $$
and therefore the "coefficient" (actually a polynomial in $u$) of $z^n$ in $exp((u-1)z)/(1-z)$ is
$$ P_n(u) = 1 + (u-1) + {(u-1)^2 \over 2!} + \cdots + {(u-1)^n \over n!} = \sum_{j=0}^n {{(u-1)^j } \over j!} $$
since division of a generating function by $1-z$ has the effect of taking partial sums of the coefficients.
The coefficient of $u^k$ in $P_n(u)$ (which I'll denote $[u^k] P_n(u)$, where $[u^k]$ denotes taking the $u^k$-coefficient) is then
$$ [u^k] P_n(u) = \sum_{j=0}^n [u^k] {(u-1)^j \over j!} $$
But we only need to do the sum for $j = k, \ldots, n$; the lower terms are zero, since they are the $u^k$-coefficient of a polynomial of degree less than $k$. So
$$ [u^k] P_n(u) = \sum_{j=k}^n [u^k] {(u-1)^j \over j!} $$
and by the binomial theorem,
$$ [u^k] P_n(u) = \sum_{j=k}^n {(-1)^{j-k} \over k! (j-k)!} $$
Finally, $F(k,n) = n! [u^k] P_n(u)$, and so we have
$$ F(k,n) = n! \sum_{j=k}^n {(-1)^{j-k} \over k!(j-k)!} $$
