How am I supposed to know that $\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty } \binom{n+1}{n}x^n$? I'm currently reading through the solution to a problem that involves finding generating functions. In some of the intermediary steps, it is written that
$$\frac{1}{(1-x)^2} = \sum_{n=0}^{\infty } \binom{n+1}{n}x^n$$
without any justification.
Is there some kind of reasoning behind this or is this just one of those things I have to accept?
 A: I'm not entirely sure why they choose to write it that way.
$$G(x) = \frac{1}{1-x} = \sum\limits_{n=0}^\infty x^n$$
$$G'(x) = \frac{1}{(1-x)^2} = \sum\limits_{n=0}^\infty nx^{n-1} = \sum_{n=0}^{\infty } \binom{n+1}{n}x^n$$
This is a minor reindexing, after noting that
$$\binom{n+1}{n} = \binom{n+1}{1} = n+1$$
It appears have been written in a deliberately obtuse manner, though perhaps it facilitates the application of Pascal's identity or similar later on?
A: Keep in mind that in general for $\alpha\in \mathbb{R}$ you have $$(1+x)^\alpha = \sum_{n=0}^{\infty} {\alpha \choose n}x^n$$ by Taylor series.  
A: Do you know the formula $ \displaystyle \frac{1}{1-x} = \sum_{n=0}^\infty x^n$, $|x| < 1$?
If you differentiate term-by-term and re-index you will get the formula you are looking for.
A: Here are some useful evaluations:
$$
1+x+x^2+\cdots+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1. \tag1
$$ Then by differentiating $(1)$ you get
$$
1+2x+3x^2+\cdots+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-nx^{n}}{1-x}, \quad |x|<1, \tag2
$$ and by making $n \to +\infty$ in $(2)$  using $|x|<1$, gives 
$$
1+2x+3x^2+\cdots+nx^{n-1}+\cdots=\frac{1}{(1-x)^2} \tag3
$$ and observe that
$$
\binom{n+1}{n}=n+1.
$$
A: $$\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots\,\,\,\,\,\forall|x|\lt1$$ Then $$\dfrac{1}{(1-x)^2}=(1+x+x^2+x^3+\cdots)(1+x+x^2+x^3+\cdots)$$ $x^k$ in the expansion of right hand side is produced by $$1.x^k+x.x^{k-1}+x^2.x^{k-2}+\cdots+x^k.1=(k+1)x^k.$$ Hence $$\dfrac{1}{(1-x)^2}=1+2x+3x^2+4x^3+\cdots.$$ Similarly we can find $\dfrac{1}{(1-x)^3}$ and so on. 
A: We have 
$$\frac{1}{(1 - x)^2} = \frac{d}{dx}\frac{1}{1 - x} = \frac{d}{dx}\sum_{n=0}^\infty x^n = \sum_{n = 0}^\infty \frac{d}{dx}x^n $$ $$= \sum_{n = 1}^\infty nx^{n-1} = \sum_{n = 0}^\infty (n+1)x^n = \sum_{n = 0}^\infty \binom{n+1}{n}x^n.$$
