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.

The sequence $y=(1,2,3,4,\ldots)$ is an output of the linear system
$$
y_{k+1}=y_k+u_k,\qquad y_0=1
$$
for the input $u=(1,1,1,1,\ldots)$. Perform the $Z$-transform (multiply by $z^k$ and add up for all $k$)
$$
\sum_{k=0}^\infty y_{k+1}z^k=\underbrace{\sum_{k=0}^\infty y_kz^k}_{y(z)}+\underbrace{\sum_{k=0}^\infty u_kz^k}_{u(z)}\quad\Rightarrow\quad
\frac{1}{z}(y(z)-y_0)=y(z)+u(z)\quad\Rightarrow
$$
$$
\Rightarrow\quad y(z)=\frac{1+z u(z)}{1-z}=\frac{1+z\frac{1}{1-z}}{1-z}=\frac{1}{(1-z)^2}.
$$
