Prove that $\left({1 \over 1-x}\right)^2 = \sum_{n=0}^\infty (n+1)x^n$ 
Prove that if $x\in [0, 1)$, then $\displaystyle \left({1 \over 1-x}\right)^2 = \sum_{n=0}^\infty (n+1)x^n$

I have a theorem which claims that

THEOREM: If for some $x\in\mathbb R$ the power series $\sum_{n=0}^\infty a_nx^n$ and $\sum_{n=0}^\infty b_nx^n$ are absolutely convergent, then $$\left(\sum_{n=0}^\infty a_nx^n\right)\left(\sum_{n=0}^\infty b_nx^n\right) = \left(\sum_{n=0}^\infty c_nx^n\right),$$ where $c_n = \sum_{k=0}^na_kb_{n-k}.$

I attempted the proof and the heart of the proof is extremely straightforward: $$\begin{align}\left({1 \over 1-x}\right)^2 & = \left({1 \over 1-x}\right)\left({1 \over 1-x}\right)\\ & = \left(\sum_{n=0}^\infty1x^n\right)\left(\sum_{n=0}^\infty1x^n\right) \tag{Geometric Series}\\ &= \sum_{n=0}^\infty\left(\left[\sum_{k=0}^n(1)(1)\right]x^n\right) \tag{Theorem}\\ &=\sum_{n=0}^\infty\left[(n+1)x^n\right].\end{align}$$
However, is it necessary for the proof to show that $\sum_{n=0}^\infty x^n$ is absolutely convergent? I'm not entirely sure how to go about doing that. 
If I had to hazard a guess, I would say that since $x \in [0, 1)$, then $|x^n| = |x|^n = x^n$ for all $n\in\mathbb N$, so $\sum_{n=0}^\infty |x^n| = \sum_{n=0}^\infty x^n$ is convergent since it is a geometric series which implies the series is absolutely convergent. But is it valid to say two infinite series are "equal" like this?
 A: Although you have the multiplication theorem at hand, but you can easily prove the problem by differentiation. That is, try to differentiate both sides of the following identity $$\frac1{1-x}=\sum_{n=0}^\infty{x^n}$$ and you'll get this $$\left(\frac1{1-x}\right)^2=\sum_{n=0}^\infty{nx^{n-1}}$$ Since the first term of the summation is zero, you can change the index to get this $$\left(\frac1{1-x}\right)^2=\sum_{n=0}^\infty{(n+1)x^{n}}$$
A: If you are allowed to start from the right, the problem is simple since $$\sum_{n=0}^\infty (n+1)x^n=\sum_{n=0}^\infty nx^n+\sum_{n=0}^\infty x^n=x\sum_{n=0}^\infty nx^{n-1}+\sum_{n=0}^\infty x^n=x \left(\sum_{n=0}^\infty x^n \right)'+\sum_{n=0}^\infty x^n$$
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
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\begin{align}
\pars{1 \over 1 - x}^{2} & = \pars{1 - x}^{-2} =
\sum_{n = 0}^{\infty}{-2 \choose n}\pars{-x}^{n} =
\sum_{n = 0}^{\infty}\braces{{-\bracks{-2} + n - 1 \choose n}\pars{-1}^{n}}\pars{-x}^{n}
\\[5mm] & = 
\sum_{n = 0}^{\infty}{n + 1 \choose n}x^{n} =
\bbx{\ds{\sum_{n = 0}^{\infty}\pars{n + 1}x^{n}}}
\end{align}
A: Your proof is correct. The geometric series converges absolutey for $|x|<1$.
So, what is your problem ?
