The most straightforward way to do this is to just do the division.
It's hard to show the process of long division on here.
\(\dfrac{1}{(1+x)^2}=\dfrac{1}{1+2x+x^2}\)
and just set this up as a usual long division, if you go through it a few steps you'll see it's
\(\dfrac{1}{1+2x+x^2} = 1 - 2x + 3x^2 - 4x^3 + 5x^4 \dots + (-1)^k (k+1)x^k,~\forall k \geq 0\)
hrm.. there might be another way to approach it that's pretty straightforward
\(\sum \limits_{k=0}^\infty ~x^k = \dfrac{1}{1-x}\)
\(\dfrac{1}{1+x}=\dfrac{1}{1-(-x)}\)
\(\dfrac{1}{1-(-x)} = \sum \limits_{k=0}^\infty ~(-1)^k x^k\)
\(\dfrac{d}{dx}~\dfrac{1}{1+x} = - \dfrac{1}{(1+x)^2}\)
\(\dfrac{1}{(1+x)^2} = -\dfrac{d}{dx} \left(\sum \limits_{k=0}^\infty~(-1)^k x^k\right) = \\ -\left(\sum \limits_{k=1}^\infty~(-1)^k k x^{k-1}\right) = \sum \limits_{k=0}^\infty~(-1)^k(k+1)x^k\)
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