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?

Let say you have the product of two power series $\mathrm A(x) \mathrm B(x)$ then you have
$$(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots)(b_0+b_1x+b_2x^2+b_3x^3+b_4x^4+\cdots)$$
The constant term is given by $a_0b_0$. The $x$-term is $(a_0b_1+a_1b_0)x$. The $x^2$-terms is
$$(a_0b_2+a_1b_1+a_2b_0)x^2$$
The $x^3$-term will be $(a_0b_3+a_1b_2+a_2b_1+a_3b_0)x^3$. The $x^k$-term is given by
$$(a_0b_k+a_1b_{k-1}+a_2b_{k-2}+\cdots+a_{k-2}b_2+a_{k-1}b_1+a_kb_0)x^k$$
$$=\left( \sum_{i=0}^k a_ib_{k-i}\right)x^k$$
In your case, where $\mathrm A(x) = \mathrm e^x$ and so $a_i = \frac{1}{i!}$, and $\mathrm B(x) = \frac{1}{1-x}$ where $b_j = 1$.
The $x^k$-term in the product of their power series will be
$$\left(\sum_{i=0}^k \frac{1}{i!}\cdot 1\right)x^k = \sum_{i=0}^k \frac{x^k}{i!}$$
This gives us the power series. Recall that the Taylor series of a function is
$$\sum_{k=0}^{\infty} \frac{\mathrm f^{(k)}(0)}{k!}\,x^k$$
Equating the $x^k$-coefficients gives
$$\sum_{i=0}^k \frac{1}{i!}=\frac{\mathrm f^{(k)}(0)}{k!} \ \ \implies  \ \ \mathrm f^{(k)}(0) \ \ = \ \ \sum_{i=0}^k \frac{k!}{i!}$$
$$\mathrm f^{(k)}(0) = \frac{k!}{0!} + \frac{k!}{1!} + \frac{k!}{2!} 
+ \cdots + \frac{k!}{(k-2)!} + \frac{k!}{(k-1)!} + \frac{k!}{k!}$$
I'm not able to simplify this any further. Maple suggests the incomplete Gamma function 
$$\mathrm e \Gamma(k+1,1) \ \ = \ \ \mathrm e \int_1^{\infty} t^k\mathrm e^{-t}~\mathrm dt$$
I have no idea what this means!
