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?

If an exponential generating function (egf) $F$ counts objects in some combinatorial class $\mathcal{F}$, then $e^F$ counts ways to construct objects with parts in $\mathcal{F}$ with relabeling. So, what does the egf $F$ count? $F$ counts cycles on one and two elements (there is one of each kind), so $e^F$ counts involutions. 
Now let property $P_i$, $1\le i\le n$, be that $i$ is in a $1$-cycle (a fixed point). Then by the Inclusion-Exclusion Principle the expression you want to simplify is exactly the number of objects that have none of these properties, i.e. we are counting involutions with no $1$-cycles, i.e. with just the $2$-cycles. Thus, the generating function for those is $e^{x^2/2}$, from which you can extract the coefficient at $x^{2n}$.
