I need help with a question, I need to show that the derangement numbers $D_n$ satisfy $$\sum_{n=0}^{+\infty} D_n \cdot \frac{x^n}{n!} = \frac{e^{-x}}{1-x}$$
I also need to prove that the expected number of fixed points of a permutation is $1$. I am so lost any help is appreciated.
On
For derangements, we get the combinatorial class
$$\def\textsc#1{\dosc#1\csod} \def\dosc#1#2\csod{{\rm #1{\small #2}}} \textsc{SET}(\textsc{CYC}_{=2}(\mathcal{Z}) + \textsc{CYC}_{=3}(\mathcal{Z}) + \textsc{CYC}_{=4}(\mathcal{Z}) + \cdots)$$
which yields the EGF
$$F(z) = \exp\left(\frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+\cdots\right)$$
which is
$$\exp\left(-z + \log\frac{1}{1-z} \right) = \frac{\exp(-z)}{1-z}.$$
For the expected number of fixed points we use
$$\textsc{SET}(\mathcal{U}\times \textsc{CYC}_{=1}(\mathcal{Z}) + \textsc{CYC}_{=2}(\mathcal{Z}) + \textsc{CYC}_{=3}(\mathcal{Z}) + \textsc{CYC}_{=4}(\mathcal{Z}) + \cdots)$$
with EGF
$$G(z, u) = \exp\left(uz + \frac{z^2}{2}+\frac{z^3}{3}+\frac{z^4}{4}+\cdots\right)$$
which is
$$\exp\left(uz - z + \log\frac{1}{1-z} \right) = \frac{\exp(uz)\exp(-z)}{1-z}.$$
We get for the expectation of the number of fixed points
$$[z^n] \left. \frac{\partial}{\partial u} G(z, u) \right|_{u=1} \\ = [z^n] \left. z\frac{\exp(uz)\exp(-z)}{1-z} \right|_{u=1} = [z^n] \frac{z}{1-z} = 1.$$
The integral representation of the derangements $D_n$ is $$D_n=\int_{0}^{\infty} (t-1)^n e^{-t} dt.$$ Then $$S=\sum_{n=0}^{\infty} \int_{0}^{\infty} \frac{x^n}{n!}(t-1)^n e^{-t} dt=\int_{0}^{\infty} e^{x(t-1)} e^{-t} dt= e^{-x}\int_{0}^{\infty} e^{-(1-x)t} dt= \frac{e^{-x}}{1-x}.$$