Converting a power series to a function 
$$\sum (n+1)^2x^n = {1+x\over (1-x)^3}$$

I tried to find a similar power series and start from there. I see $ x^n $ and it reminds me of geometric series, but $(n+1)^2$ confuses me. How do I prove this?

Just another way to do it.
$$A=\sum_{n=0}^\infty (n+1)^2x^n =\sum_{n=0}^\infty (n^2+2n+1)x^n $$ Now, write $$n^2+2n+1=(n(n-1)+n)+2n+1=n(n-1)+3n+1$$ $$A=\sum_{n=0}^\infty n(n-1)x^n+3\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n$$ $$A=x^2\sum_{n=0}^\infty n(n-1)x^{n-2}+3x\sum_{n=0}^\infty nx^{n-1}+\sum_{n=0}^\infty x^n$$ $$A=x^2\left(\sum_{n=0}^\infty x^n \right)''+3x\left(\sum_{n=0}^\infty x^n \right)'+\left(\sum_{n=0}^\infty x^n \right)$$ $$\left(\sum_{n=0}^\infty x^n \right)=\frac{1}{1-x}\implies\left(\sum_{n=0}^\infty x^n \right)'=\frac{1}{(1-x)^2}\implies\left(\sum_{n=0}^\infty x^n \right)''=\frac{2}{(1-x)^3}$$ $$A=\frac{2x^2}{(1-x)^3}+\frac{3x}{(1-x)^2}+\frac{1}{1-x}=\frac{1+x}{(1-x)^3}$$
Using the same steps, you could make the problem more general and show that $$B=\sum_{n=0}^\infty (an^2+bn+c)x^n =\frac{(a-b+c)x^2 + (a+b-2 c)x+c }{(1-x)^3}$$
