<s>[INST] 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?
 [/INST] $$\dfrac{1}{1-x}=1+x+x^2+x^3+\cdots\,\,\,\,\,\forall|x|\lt1$$ 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. 
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