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 

$$h^{(n)}(x) = \sum_{k=0}^n {n \choose k} (n-k)! = \sum_{k=0}^n \frac{n!}{k!} = n! \sum_{k=0}^n \frac{1}{k!}$$
Let $$y=\frac{e^x}{1-x},$$ or

$$y(1-x)=e^x.$$

By differentiation,

$$y'(1-x)-y=e^x,\\
y''(1-x)-2y'=e^x,\\
y'''(1-x)-3y''=e^x,\\
y''''(1-x)-4y'''=e^x,\\
\cdots$$

and obviously

$$y^{(n)}(1-x)-ny^{(n-1)}=e^x.$$

Evaluating at $0$, we get the recurrence

$$y_n=ny_{n-1}+1$$ with $y_0=1$.



To solve it, let $z_n:=\dfrac{y_n}{n!}$. (By the way, the $z_n$ are precisely the Taylor coefficients.) We have

$$n!z_n=n(n-1)!z_{n-1}+1$$ or

$$z_n=z_{n-1}+\frac1{n!}.$$

Hence,

$$\frac{y_n}{n!}=z_n=\sum_{k=0}^n\frac1{k!}.$$