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.

Yes, first idea is right. More formally,
$$
\left(\sum_{i=0}^\infty (-1)^i x^i \right)
\left(\sum_{j=0}^\infty (-1)^j x^j \right)
= \sum_{n=0}^\infty \sum_{i=0}^n (-1)^i x^i (-1)^{n-i}x^{n-i}
= \sum_{n=0}^\infty \sum_{i=0}^n (-1)^n x^n
= \sum_{n=0}^\infty (n+1) (-1)^n x^n.
$$
On the second one, note that
$$
\frac{d}{dx} (1+x)^{-1} = -(1+x)^{-2},
$$
thus
$$
(1+x)^{-2}
 = -\frac{d}{dx} (1+x)^{-1}
 = -\frac{d}{dx} \sum_{i=0}^\infty (-1)^i x^i
 = -\sum_{i=0}^\infty (-1)^i \frac{d}{dx} x^i
 = \sum_{i=1}^\infty (-1)^{i-1}ix^{i-1}
 = \sum_{i=0}^\infty (-1)^i (i+1)x^i
$$
