Verifying Newton's Binomial Series for $(1+x)^{−2}$ 
Verify the consistency of Newton’s Binomial series for the function
  $(1+x)^{−2}$ in two ways: 
$(a)$ by multiplying the usual geometric series for $(1 +x)^{−1}$with
  itself 
$(b)$ by differentiation of the series for $(1 +x)^{−1}$

$(a)$ I have that Newton's Binomial series says that
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
I have that the geometric series for $(1 +x)^{−1}$ is given by
$$\begin{align*}
\frac{1}{1+x}
&=\sum_{k=0}^{\infty}(-x)^k\\\\
&=-x^0+(-x)^1+(-x)^2+(-x)^3+...\\\\
&=1-x+x^2-x^3+...
\end{align*}$$
Thus
$$\begin{align*}
\left(\frac{1}{1+x}\right)^2
&=(1-x+x^2-x^3+...)(1-x+x^2-x^3+...)\\\\
&=(1-x+x^2-x^3+...)\\\\
& +(-x+x^2-x^3+...)\\\\
& +(x^2-x^3+...)\\\\
& +(-x^3+...)\\\\
&=1-2x+3x^2-4x^3+...
\end{align*}$$
Checking that this equals Newton's Binomial series
$$(1+x)^p=1+px+\frac{p(p−1)}{2!}x^2+\frac{p(p−1)(p−2)}{3!}x^3+···$$
where $p=-2$
$$\begin{align*}
(1+x)^{-2}
&=1-2x+\frac{-2(-3)}{2!}x^2+\frac{-2(-3)(-4)}{3!}x^3+···\\\\
&=1-2x+3x^2-4x^3+... \checkmark
\end{align*}$$
Is this a valid proof?
$(b)$
I have that $$\int_0^x\frac{dt}{1+t}=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}$$
but I'm not sure how to use this fact.

The binomial expansion is
$$
\left( {1 + x} \right)^{\, - 2}  = \sum\limits_{0\, \le \,k} {\left( \matrix{
   - 2 \cr 
  k \cr}  \right)x^{\,k} }  = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
  k + 1 \cr 
  k \cr}  \right)x^{\,k} }  = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)x^{\,k} } 
$$
where "upper negation" of the binomial has been used
The multiplication of geometric series gives
$$
\eqalign{
  & {1 \over {1 + x}}{1 \over {1 + x}} = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} x^{\,k} } \sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,j} x^{\,j} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {\sum\limits_{0\, \le \,j} {\left( { - 1} \right)^{\,k + j} x^{\,k + j} } } \quad \left| \matrix{
  \;0 \le k + j = s \hfill \cr 
  \;0 \le j = s - k\; \Rightarrow \;0 \le k \le s \hfill \cr}  \right.\;\quad  =   \cr 
  &  = \sum\limits_{0\, \le \,s} {\left( {\sum\limits_{0 \le k \le s} 1 } \right)\left( { - 1} \right)^{\,s} x^{\,s} }  = \sum\limits_{0\, \le \,s} {\left( { - 1} \right)^{\,s} \left( {s + 1} \right)x^{\,s} }  \cr} 
$$
and differentiating the geometric series
$$
\eqalign{
  & {d \over {dx}}\left( {1 + x} \right)^{\, - 1}  =  - \left( {1 + x} \right)^{\, - 2}  = {d \over {dx}}\sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} x^{\,k} }  =   \cr 
  &  = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} k\,x^{\,k - 1} }  = \sum\limits_{1\, \le \,k} {\left( { - 1} \right)^{\,k} k\,x^{\,k - 1} }  =   \cr 
  &  = \sum\limits_{1\, \le \,k} {\left( { - 1} \right)^{\,k + 1} \left( {k + 1} \right)\,x^{\,k} }  =  - \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( {k + 1} \right)x^{\,k} }  \cr} 
$$
