what is the nth-derivative in 0 of $\frac{e^x}{1-x}$ Using $$f^n(0)=n! .a_n$$
I have$$\frac{e^x}{1-x}=\sum{\left(\frac{x^n}{n!}\right)}\sum{x^n}=\sum_{n \geq 0}\left({\sum_{k=0}^{n}{\frac{n!}{k!n!}}}\right)$$
How to get the combination?

The General Leibniz rule tells us that for two smooth functions $f,g$, the $n$th derivative of $h(x) = f(x)g(x)$ is 
$$h^{(n)}(x) = \sum_{k=0}^n {n \choose k} f^{(n-k)}(x)g^{(k)}(x)$$
The $n$ derivative of $e^x$ at $x=0$ is $1$, and the $n$th derivative of $\frac{1}{1-x}$ at $x=0$ is $n!$ by considering the power series. Hence, the $n$th derivative of $h(x) = \frac{e^x}{1-x}$ at $x=0$ is 
$$h^{(n)}(x) = \sum_{k=0}^n {n \choose k} (n-k)! = \sum_{k=0}^n \frac{n!}{k!} = n! \sum_{k=0}^n \frac{1}{k!}$$
