Integral transform and $\frac{1}{n!}$. Probably this is a trivial question, but I am unable to find an answer: is there a function $v(x)$ such that 
$$
\int_{0}^\infty x^n e^{v(x)} dx =\frac{1}{n!}
$$
for all positiv integer n?
 A: Such a function would have to satisfy:
$$\int(1-x)^2e^{v(x)}dx=-\frac12,$$
but the left hand side is clearly non-negative...
EDIT: Another contradiction for positive exponents is
$$\int x(1-x)^4e^{v(x)}dx = -\frac{19}{120}$$
A: Your question is a special case of the Hamburger moment problem: given a sequence of positive numbers $(\mu_n)_{n\geq 0}$  decides if there exists a  positive measure $\mu$ on $\mathbb{R}$ such that $\newcommand{\bR}{\mathbb{R}}$
$$\mu_n=\int_{\bR} x^n \mu(|dx|),\;\;\forall n=0,1,2,\dotsc. $$
There exist many necessary  and sufficient  conditions for the existence of such a measure. The classical text  The Laplace Transform by D.V. Widder is a good place to look, especially Section 10-13 of Chapter III. Here is a link to the book.
Addendum 1.  If you do not care about the positivity of the measure $\mu$ then we have the following result of  R.P. Boas. (See Section 14, Chapter III of Widder's book)
For any sequence of real numbers $(\mu_n)_{n\geq 0}$ there exists a signed measure $\mu$ on $[0,\infty)$ with the following properties
$$ \mu_n=\int_0^\infty x^n \mu(dx) ,\;\;\forall n=0,1,2,\dotsc, \tag{1} $$
and 
$$\int_0^\infty|\mu|(dx)<\infty.\tag{2} $$
Recall that any signed measure $\mu$ is the difference of two positive measures $\mu=\mu^+-\mu^-$ and the total variation  mesure is $|\mu|=\mu^++\mu^-$. On the semiaxis $[0,\infty)$ a signed measure $\mu$  has the from $\mu=d\alpha$, where $\alpha$ is a function of   bounded variation on compact intervals.
Addendum 2.    Consider the Fourier transform  of the  measure $\mu$ in Addendum 1. We have $\newcommand{\ii}{\boldsymbol{i}}$
$$ \widehat{\mu}(\xi):=\int_{\bR} e^{-\ii\xi x} \mu(dx) =\sum_{n\geq 0}\mu_n\frac{(-\ii\xi)^n}{n!}.\tag{3} $$
There is an obvious problem   with the above  equality: the series in the right-hand side may not be convergent for all $\xi$ if the momenta  $\mu_n$ grow too fast.    In fact  if the momenta grow fast, there  exist at least  two measures $\mu$, $\mu'$ satisfying both (1) and (2) above. In  your case  the $\mu_n$   decay very fast and my guess is that $\mu$ is unique.  (The  series  in the right-hand side of (3)   converges for any $\xi$ so it defines  a continuous function  $f(\xi)$ which can be viewed as the Fourier transform of $\mu$ in the sense of distributions. Now  use the Fourier inversion formula to recover $\mu$.)
In any case, the space of solutions of (1) (2) is completely understood. A good place to look is Akhiezer  book The Classical Moment problem or  Chapter 16 in the  book Unbounded selfadjoint operators on Hilbert space by Konrad Schmudgen. The story is quite rich.
A: No. For the integral to make sense, $v$ must be at least measurable. Let 
$$X_k=\{x\in [1,\infty): v(x)>-k\}$$
and note that $[1,\infty)=\cup_{k\in\mathbb N} X_k$ so some set $X_k$ has positive measure by countable subadditivity. Then
$$\int_0^\infty x^n e^{v(x)}dx\ge \int_{X_k} x^n e^{v(x)}dx\ge \int_{X_k} e^{-k}dx > \frac{1}{n!}$$
for sufficiently large $n$.
A: Suppose you are not wedded to the interval $(0,\+\infty)$.  
If $C$ is the unit circle in the complex plane, oriented in the counter-clockwise direction as usual, then
$$
\frac{1}{2\pi i}\oint_C z^n \frac{e^{1/z}}{z}\;dz = \frac{1}{n!}
$$
for $n=0,1,2,\dots$
