Closed-form for modified formal power series This question have been driving me crazy for months now. This comes from work on multiple integrals and convolutions but is phrased in terms of formal power series.
We start with a formal power series
$P(C) = \sum_{n=0}^\infty a_n C^{n+1}$
where $a_n = (-1)^n n!$
With these coefficients the formal power series can be expressed as a hypergeometric function
$P(C) = C \, _2F_0(1,1;;-C)$
I'm then interested in the formal power series $P_T(C)=\frac{P}{1-P}$ as well as, if possible, the series $P^n$ for arbitrary positive integer n (where this is the power series P raised to the nth power).
Specifically if
$P_T(C) = \sum_{n=0}^\infty b_n C^{n}$ 
then want to construct the function
$f(x) = \sum_{n=0}^\infty \frac{b_{n+1}}{(n!)^2} x^{n}$
which, from other results, should converge for all x. We can think of this as the doubly-exponential generating function for the $b_n$ sequence.
There are rules for multiplying and dividing formal power series (see here) and I've used these to get a recurrence relation for the coefficients in $P_T(C)$ (as well as P^n(C)) but I've been unable to solve these recurrence relations explicitly. They're in a form where each $b_n$ depends on all the previous $b_n$'s and I've not been able to make progress with them.
Explicitly the recurrence relation for the $b_n$ is $b_0 = 1$, $b_n = \sum_{k=1}^n b_{n-k} a_k$ (for n > 1). This looks simple enough but I don't think has a nice closed-form expression.
Nevertheless I do know what the $b_n$ are. They are the sequence A052186 (up to plus and minus signs). So 
$P_T(C) = C+C^3-3 C^4+14 C^5-77 C^6+497 C^7-3676 C^8+\ldots$
and
$f(x) = 1 + \frac{1}{4}x^2 - \frac{1}{12}x^3 + \frac{7}{288}x^4 - \frac{77}{14400} x^5 + 
\frac{497}{518400}x^6 +\ldots$
The question is, is it possible to figure out the function $f(x)$!? Does it have a nice closed form? Perhaps in the form of a hypergeometric function? If it does, great, if it doesn't then at least I can stop searching for it!
 A: Of course your series $P(C)$ diverges.  But it is a transseries.  Or an asymptotic series.  In fact, one of the best known.  The series
$$
\sum_{n=0}^\infty (-1)^n n! C^{n+1}
$$
is the asymptotic series (as $C \downarrow 0$) for the function
$$
p(C) = -e^{1/C} \mathrm{Ei}(-1/C) .
$$
So, of course, your series $P_T(C)$ is the asymptotic series for
$$
p_T(C) = -1 + \frac{1}{1+e^{1/C} \mathrm{Ei}(-1/C)} .
$$
Now all we need is to remember the formal relation between an ordinary generating function and an exponential generating function, and apply it twice to $p_T$.  Maybe that is not so easy?
A: [update]  remark: I shifted the original answer to the bottom although it was missing the question to keep it as reference.
I've done a nice table, which shows the coefficients of a powerseries with constant term =1; for any p 'th integer power of the function you insert p into the formula.
Unfortunately the Latex-notation of the table was too messy, so I insert it as an image here:    

(source: helms-net.de) 
If you have Pari/GP you can easily manipulate power series; the formula $ \small P \cdot (1 - P)^{-1}$, where P is defined to be a powerseries can directly be evaluated :     
P = Ser( sum(k=0,20, (-1)^k*k! * x * x^k ))
print ( P*(1 - P)^-1 + O(x^16) )

The matrix-operator is simply the coefficients of each power of the taylor series of a function in a column of an (ideally infinite sized) matrix.

[original answer]
Assumed that I got your function correct ( $ \small  P(x) = \sum_{k=1}^{\infty} (-1)^{k-1} \cdot (k-1)! \cdot x^k $ ) I have polynomials for the coefficients of $ \small P^{\text{ o } h} (x) $ Here I assume, that your notation $ \small P^n$  means iteration and not power. (If that was wrong I can delete this answer) I constructed the matrix-logarithm of the matrix-operator $ \small M$ for the function $ \small P(x)$ and got by $ \small \exp(h \cdot \log(M)) $ the following powerseries, where the coefficients at x are polynomials in the iteration-parameter h :    

$ \qquad \small  \begin{array} {l} P^{\text{ o } h} (x) = & 1 \cdot x \\\
 & - h \cdot x^2 \\\ 
 & + (h^2 + h) \cdot x^3 \\\
 & + (-h^3 - 5/2 \cdot h^2 - 5/2 \cdot h) \cdot x^4 \\\ 
 & + ( h^4 + 13/3 \cdot h^3 + 9 \cdot h^2 + 29/3 \cdot h) \cdot x^5 \\\ 
 & + (-h^5 - 77/12 \cdot h^4 - 125/6 \cdot h^3 - 511/12 \cdot h^2 - 295/6 \cdot h) \cdot x^6 \\\
 & + O(x^8)  \end{array} $ 
I was not yet able to construct the other function $ \small P_T(C)$ although I've tried with a certain vague idea; please show some of the coefficents $ \small b_n $ so that I can compare.
A: I post this as an answer since this way it's more easy to read. But it's a successive comment to my comment above. 
The formula I mean is the following: Let 
$$f(x) = 1 + \sum_{n >0}b_nx^n$$ be a formal power series over a commutative ring with unit. Then $$1/f = 1 + \sum_{n>0}c_nx^n,$$ 
$$c_n = \sum_{k=1}^n(-1)^k \sum b_{i_1}\cdots b_{i_k}$$ where the inner sum is taken over all tuples $(i_1, \dots, i_k) \in \lbrace 1, \dots, n \rbrace ^k$ such that $i_1 + \dots + i_k = n$. 
Applied to $1-P$ this yields $1/(1-P) = 1 + \sum_{n>0}c_nx^n$ with 
$$c_n = (-1)^n\sum_{k=1}^n(-1)^k \sum (i_1-1)! \cdots (i_k-1)!$$
But actually I don't believe that this is very helpful at all. 
A: I stumbled across a similar question on math overflow here. That had lot of insights into P including a simple continued fraction expression for P. Details in a paper here.
They give
$P(x) = [1, x, x, 2x, 2x, 3x, 3x, . . . , nx, nx, . . .]$
in continued fraction notation.
This surely means that
$P_T(x)+1 = \frac{1}{1-P(x)}= [1, 1, x, x, 2x, 2x, 3x, 3x, . . . , nx, nx, . . .]$
Perhaps written like this maybe it is simple enough to have a nice solution. Maybe?
