New Idea to prove $1+2x+3x^2+\cdots=(1-x)^{-2}$ 
Given $|x|<1 $ prove that $\\1+2x+3x^2+4x^3+5x^4+...=\frac{1}{(1-x)^2}$.

1st Proof: Let $s$ be defined as
$$
s=1+2x+3x^2+4x^3+5x^4+\cdots
$$
Then we have
$$
\begin{align}
xs&=x+2x^2+3x^3+4x^4+5x^5+\cdots\\
s-xs&=1+(2x-x)+(3x^2-2x^2)+\cdots\\
s-xs&=1+x+x^2+x^3+\cdots\\
s-xs&=\frac{1}{1-x}\\
s(1-x)&=\frac{1}{1-x}\\
s&= \frac{1}{(1-x)^2}
\end{align}
$$
2nd proof:
$$
\begin{align}
s&=1+2x+3x^2+4x^3+5x^4+\cdots\\
&=\left(1+x+x^2+x^3+\cdots\right)'\\
&=\left(\frac{1}{1-x}\right)'\\
&=\frac{0-(-1)}{(1-x)^2}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
3rd Proof:
$$
\begin{align}
s=&1+2x+3x^2+4x^3+5x^4+\cdots\\
=&1+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+x+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+x^2+x^3+x^4+x^5+\cdots\\
&+0+0+0+x^3+x^4+x^5+\cdots\\
&+\cdots
\end{align}
$$
$$
\begin{align}
s&=\frac{1}{1-x}+\frac{x}{1-x}+\frac{x^2}{1-x}+\frac{x^3}{1-x}+\cdots\\
&=\frac{1+x+x^2+x^3+x^4+x^5+...}{1-x}\\
&=\frac{\frac{1}{1-x}}{1-x}\\
&=\frac{1}{(1-x)^2}
\end{align}
$$
These are my three proofs to date. I'm looking for more ways to prove the statement.

Use differences of a sum.

$$
\begin{array}{r}
S &=& +1 & +2 x & +3 x^2 & +4 x^3 & +5 x^4 & +6 x^5 & +7 x^6 & \cdots\\
-2 x S &=& & -2x & -4 x & -6 x^2 & -8 x^3 & -10 x^4 & -12 x^5 & \cdots\\
x^2 S &=& & & +x^2 & +2 x^2 & +3 x^4 & +4 x^5 & +5 x^6 & \cdots\\
&&&&&&&&&& +\\
\hline\\
\big( 1 - 2 x + x^2 \big) S &=& +1
\end{array}
$$

Or

$$
\begin{array}{rclc}
S &=& \displaystyle + \sum_{k=1}^\infty k x^{k-1}\\
- 2 x S &=& \displaystyle - \sum_{k=1}^\infty 2 k x^{k}\\
x^2 S &=& \displaystyle + \sum_{k=1}^\infty k x^{k+1}\\
&&&+\\
\hline\\
\big( 1 - 2 x + x^2 \big) S
&=& \displaystyle \sum_{k=1}^\infty k x^{k-1}
  - \sum_{k=1}^\infty 2 k x^{k}
  + \sum_{k=1}^\infty k x^{k+1}\\
&=& \displaystyle \sum_{k=0}^\infty (k+1) x^k
  - \sum_{k=1}^\infty 2 k x^{k}
  + \sum_{k=2}^\infty (k-1) x^k\\
&=& \displaystyle 1 + \big[ 2 - 2 \big] x
  + \sum_{k=2}^\infty \big[ (k+1) - 2 k + (k-1) \big] x^k\\
&=& 1
\end{array}
$$

So

$$
S = \frac{1}{1-2x+x^2} = \frac{1}{(1-x)^2}
$$

