<s>[INST] 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.
 [/INST] Here is another variation.
Assuming the geometric series $\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}$ is known, we consider functions $f,g:(-1,1) \rightarrow \mathbb{R}$
\begin{align*}
f(x)&=\sum_{n=0}^{\infty}(n+1)x^n\qquad\qquad g(x)=\frac{1}{(1-x)^2}
\end{align*}

We obtain
  \begin{align*}
\int f(x) dx&=\int\left(\sum_{n=0}^{\infty}(n+1)x^n\right) dx
=\sum_{n=0}^{\infty}\int(n+1)x^n dx\\
&=\sum_{n=0}^{\infty}x^{n+1}+C=\sum_{n=1}^\infty x^n+C\\
&=\frac{1}{1-x}-1+C\\
\\
\int g(x) dx&= \int \frac{1}{(1-x)²}dx=\frac{1}{1-x}+D
\end{align*}
We observe $f$ and $g$ have the same antiderivative $\frac{1}{1-x}$ differing by a constant only. Since $f(0)=g(0)=1$ they are equal.

</s>