How to prove this Taylor expansion of $\frac{1}{(1+x)^2}=-1\times\displaystyle\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$? I came across this series of the Taylor Expansion-

$$\frac{1}{(1+x)^2}=1 - 2x + 3x^2 -4x^3 + \dots.=-1\times\sum_{n=1}^{\infty}(-1)^nnx^{n-1}$$

But I have no idea how to prove this...
Thanks for any help!

Observe that 
$$\frac x{(1+x)^2}=x - 2x^2 + 3x^3 -4x^4 + \dots$$ and adding the original
$$\frac1{(1+x)^2}=1-2x+3x^2-4x^3+5x^4\cdots$$
you verify
$$\frac x{(1+x)^2}+\frac1{(1+x)^2}=1-x+x^2-x^3+x^4-\cdots=\frac1{x+1}.$$
