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?

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 1013 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 righthand 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 righthand 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. 


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 $n$. 


Such a function would have to satisfy: $$\int(1x)^2e^{v(x)}dx=\frac12,$$ but the left hand side is clearly nonnegative... EDIT: Another contradiction for positive exponents is $$\int x(1x)^4e^{v(x)}dx = \frac{19}{120}$$ 


Suppose you are not wedded to the interval $(0,\+\infty)$. If $C$ is the unit circle in the complex plane, oriented in the counterclockwise direction as usual, then $$ \frac{1}{2\pi i}\oint_C z^n \frac{e^{1/z}}{z}\;dz = \frac{1}{n!} $$ for $n=0,1,2,\dots$ 

