Enumerative Combinatorics Problem I'm working on a problem from Enumerative Combinatorics by Richard Stanley.  Problem 7 Section 1 says

Let $e^{x+\frac{x^2}{2}}=\sum_{n\geq 0}f(n)\frac{x^n}{n!}$.  Find a simple expression for $\sum_{i=0}^n(-1)^{n-i}\binom{n}{i}f(i)$.

Earlier in the section $f(n)$ is defined to be $\sum_{j\geq 0}\binom{n}{2j}\frac{(2j)!}{2^j j!}$, but I suspect that we should be able to do the problem without that.
We tried using a recurrence relation, and that seemed to lead no where.  We thought we were clever and tried substituting $x+1$ in for $x$ to pick up the binomial terms, but soon found we were not actually clever.
Any suggestions of how to move forward?

Just expanding Marko Riedel's comment,
$$\begin{eqnarray*}e^{-x}\sum_{n\geq 0}f(n)\frac{x^n}{n!}&=&\left(\sum_{m\geq 0}(-1)^m\frac{x^{m}}{m!}\right)\left(\sum_{n\geq 0}f(n)\frac{x^n}{n!}\right)\\&=&\sum_{r\geq 0}x^r\sum_{h=0}^{r}\frac{(-1)^{r-h}}{(r-h)!h!}\,f(h)\\&=&\sum_{r\geq 0}\frac{x^r}{r!}\sum_{h=0}^{r}(-1)^{r-h}\binom{r}{h}\,f(h).\tag{1}\end{eqnarray*}$$
On the other hand, the LHS of $(1)$ equals:
$$ e^{x^2/2}=\sum_{p\geq 0}\frac{x^{2p}}{2^p\,p!}=\sum_{p\geq 0}\frac{x^{2q}}{(2p)!}\cdot\frac{(2p)!}{2^p\,p!}\tag{2}$$
so your sum equals zero if $n$ is odd and $\large\frac{n!}{2^{n/2}(n/2)!}$ otherwise.
