Simplifying a Taylor polynomial that involves Stirling numbers of the second kind I am currently trying to evaluate a complicated function $f$ at a point $x+a$ using a high order Taylor polynomial about the point $a$. The polynomial has the standard form:
$$\sum_{n=1}^k \frac{x^n}{n!} f^{(n)}(a)$$
However it takes a long time to evaluate because the $n^{th}$ derivative requires me to compute the value of a second summation:
$$f^{(n)}(a) =  \sum_{m=1}^n (-1)^{m-1} (m-1)!S(n,m) a^m$$
Here, $S(n,m)$ denotes a Stirling number of the second kind and the integer sequence $(m-1)!S(n,m)$ is a well-documented integer sequence (A028246).
My question is as follows: it possible to simply this Taylor polynomial i.e.,
$$ \sum_{n=1}^k \frac{x^n}{n!} \sum_{m=1}^n (-1)^{m-1} (m-1)!S(n,m) a^m$$
into a single sum that has the form:
$$ \sum_{n=1}^k C_n x^n a^n$$
Here the $C_n$ are constants that do not depend on $x$ or $n$ (and could have closed form expressions or be computed using recursion).
 A: Changing the order of summation, we get
$$ \sum_{n=1}^k \frac{x^n}{n!} \sum_{m=1}^n (-1)^{m-1} (m-1)!S(n,m) a^m$$ 
$$ = \sum_{m=1}^k (-1)^{m-1} (m-1)! a^m \sum_{n=m}^k \frac{x^n}{n!} S(n,m).$$
Using formula (13) from http://mathworld.wolfram.com/StirlingNumberoftheSecondKind.html , we further get
$$= \sum_{m=1}^k (-1)^{m-1} (m-1)! a^m\  [y^k]\ \frac{(e^{xy}-1)^m}{m!(1-y)} $$ 
$$= [y^k]\ \frac{1}{1-y}\sum_{m=1}^k (-1)^{m-1} \frac{((e^{xy}-1)a)^m}{m}.$$ 
Since $e^{xy}-1 = xy + O(y^2)$, we can extend the summation to $+\infty$ without affecting the result (the extra terms have $y$ in powers $>k$). So, the original expression equals the coefficient of $y^k$ in the generating function:
$$\frac{1}{1-y}\sum_{m=1}^{\infty} (-1)^{m-1} \frac{((e^{xy}-1)a)^m}{m}$$
$$ = \frac{\log(1+(e^{xy}-1)a)}{1-y}.$$
P.S. This also implies a closed-form expression for the infinite sum:
$$\sum_{n=1}^\infty \frac{x^n}{n!} f^{(n)}(a) = \log(1+(e^{x}-1)a).$$
A: Mathematica says that 
$$f^{(n)}(a) = (-1)^{n+1} \text{Li}_{1-n}\left(1-\frac{1}{a}\right).$$
A: I don't know if this will help, but it seems that $f^{(n)}(a)$ consists of the terms in positive powers of $a$ for the asymptotic series of $$- (n-1)!\; (\ln(1-1/a))^{n}$$ 
