Evaluate the sum $ \sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i} f(i)$ I need to evaluate the following sum, which depends on $n \in \mathbb N$ (call it $S(n)$ if you will)
$$ \sum_{i=0}^{n} (-1)^{n-i} \binom{n}{i} f(i)$$
where $f$ defined over $\mathbb N$ is determined by the identity
$$ \sum_{n \geq 0} f(n) \frac{x^n}{n!} = exp \left ( x+\frac{x^2}{2} \right)$$
This is a problem left as an exercise to the reader in Richard Stanley's "Enumerative Combinatorics", in the first few pages of Chapter 1, and I assume it should be simple but none of my approaches, including searching for identities involving binomial coefficients, have worked.
Thank you in advance!

Hint: $$\left(\displaystyle \sum_{l \geq 0} f(l) \frac{x^l}{l!}\right).\left(\sum_{m \geq 0} (-1)^m \frac{x^m}{m!} \right) = \quad .......$$
