Regarding the expression of $f(1)$
in terms of other known constants:
$$f(1) = -e\left(\gamma + \sum_{n\geq 1} (-1)^n \frac1{n\cdot n!} \right)
\qquad\quad \tag{$*$} \label{id}$$
$$\qquad\qquad\qquad= -e \left(\gamma - 1 + \frac1{4} - \frac1{18} + \frac1{96} - \frac1{600} + \cdots \right)$$
where $e = \sum_{n\geq0} \frac1{n!} \approx 2.718$ and
$\gamma \approx 0.577$ is the Euler-Mascheroni constant.

This infinite sum expression
follows from the identity $f(1) = -e {\rm Ei}(-1)$
and is on the Wikipedia page for the Gompertz constant.
I recommend also taking a look at the survey Euler's constant: Euler's work and modern developments by Jeffrey Lagarias,
where it appears as (2.5.11).

Since
$f(x) ``=" \sum_{n\geq 0} (-1)^n n! x^n$
and
$e^{-1} = \sum_{n\geq 0} (-1)^n \frac{1}{n!}$,
the identity ($*$) can be expressed more symmetrically as
$$ - \left(\sum_{n\geq 0} (-1)^n n! \right) \left(\sum_{m\geq 0} (-1)^m \frac{1}{m!} \right)
\quad ``=" \quad \gamma + \sum_{n\geq 1} (-1)^n \frac1{n\cdot n!}.$$

This formula suggests an argument for evaluating $f(1)$ which avoids differential equations, if one is willing to make some convenient cancellations of divergent series after rearranging the two-index summation on the left-hand side.

For arbitrary nonzero $x$, the relation ($*$) generalizes to $f(x) = - \frac1{x} e^{1/x} {\rm Ei}(-\frac1{x})$
where ${\rm Ei}(x)$ denotes the exponential integral
$${\rm Ei}(x) = \int_{-\infty}^x \frac{e^t}{t} dt. $$
Using the Taylor expansion of ${\rm Ei}(x)$,
$$f(x) = - \frac1{x} e^{1/x} \left(\gamma + \ln x + \sum_{n\geq 1} \frac{(-1)^n}{n \cdot n!} \frac1{ x^n} \right).$$

The American Mathematical Monthly, Vol.113#1 (Jan.2006), 57–62. $\endgroup$ – Noam D. Elkies Jan 14 '12 at 19:32