The question uses the textbook example of a series with radius of convergence 0: $$f=\sum n!x^n \qquad (1)$$ I am going to take the question as :" Is this $f$ of any use?" A use would be to prove some identity about factorials. One which comes to mind is
$$1+\sum_1^{n}(k-1)(k-1)!=n! \qquad (*)$$ To misquotes Samuel Johnson: "Proving (*) using (1) is like a dog walking on it's hinder legs. It is not done well; but you are surprised to find it done at all." Actually I think there is some value to this exercise (but if there is, I need to digress a bit and go a bit slow on a few details to derive it.) Let me first give two other brief proofs.

I like counting proofs. Consider the $n!$ permutations of the positive integers up to $n$. Except for the identity (that's 1), each has a greatest number which is not mapped to itself $m(\sigma)=\max\{j : \sigma(j) \ne j\}$ . So $2\le m(\sigma)\le n$ and the number of permutations with $m(\sigma)=k$ is $(k-1)(k-1)!$. QED

The standard proof of (*) would probably be by induction using the identity
$$n!=n\cdot (n-1)!=1\cdot(n-1)!+(n-1)(n-1)! \qquad(2)$$ for the induction step.

So what about using $f$ to prove the identity? $$x^2f'=\sum_0^{\infty}k\cdot k!x^{k+1}=\sum_1^{\infty}(k-1)(k-1)!x^k$$ and $g=\frac{1}{1-x}=\sum_0^{\infty} x^k$. So we need to show that $$f=\frac{1}{1-x}+\frac{1}{1-x}x^2f'\qquad(?)$$ (the first use of $g$ for the $+1$ and the second to sum $(k-1)(k-1)!\ $ ).

One way to do this is: $$xf=\sum_0^{\infty} n!x^{n+1}=\sum_1^{\infty} (n-1)!x^{n}$$ So differentiating directly: $$(xf)'=\sum_1^{\infty} n\cdot (n-1)!x^{n-1}=\sum_1^{\infty} n!x^{n-1} \qquad(4)$$ And hence $$1+x\cdot(xf)'=f \qquad(5)$$ But using the product rule, $$(xf)'=f+xf' \qquad(6)$$ Substituting (6) in (5), $$1+xf+x^2f'=f$$ and a little manipulation gives (?) as desired.

So how different are the three proofs? The identity (2) in the inductive proof might be justified as part of the inductive definition of $n!$ or as part of the counting proof that $|S_n|=(1+(n-1))|S_{n-1}|$ because each $\sigma \in S_n$ either fixes $n$ or does not. If you buy that, then the first proof is just the repeated application of this identity. And what did we use, if not convergence, for the generating function proof? The powers of $x$ just kept things in order. One important step was (4) where we used the identity (2).

For fancier uses, this article Notes on Euler’s work on divergent factorial series and their associated continued fractions says that Euler called the alternating form of $f$ the *divergent series par excellence* and takes it from there (There might be better articles to quote).

Cantor Expansionof an integer; so perhaps there are applications lurking .... – Suvrit Nov 12 '10 at 13:39