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!

Here isn't a VERFICATION of hte fact, but a full derivation, as if you had to come up with this question all by yourself to ask someone else.
Recall that taylor's theorem states:
Given an smooth function $f(x)$ around a point $a$ (that is the function is infinitely differentiable around a, so $f(a)$, $f'(a)$, $f''(a)$ all exist. 
Then it is the case that around this point
$$ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3 + ... $$
So given:
$$ f(x) = \frac{1}{(1+x)^2}$$ 
Observe that 
$$ f'(x) = -\frac{2}{(1+x)^3} $$
$$ f''(x) = \frac{2 \times 3}{(1+x)^4}$$
$$ f'''(x) = -\frac{2 \times 3 \times 4}{(1+x)^4}$$
And catching a pattern we prove in general that if:
$$ f^{(n)}(x) = (-1)^n \frac{(n+1)!}{(1+x)^{n+2}}$$
Then we verify by differentiating this that:
$$ f^{(n+1)}(x) = (-1)^{n+1} \frac{(n+2)!}{(1+x)^{n+3}}$$
So now we derive the series from taylors formula, 
$$ f(x) = f(a) + f'(a)(x-a) + \frac{1}{2!}f''(a)(x-a)^2 + \frac{1}{3!}f'''(a)(x-a)^3 + ... $$
Letting $a = 0$
$$ f(x) = f(a) + f'(a)x + \frac{1}{2!}f''(a)x^2 + \frac{1}{3!}f'''(a)x^3 + ... $$
Now substituting:
$$ \frac{1}{(1+x)^2} = 1-\frac{2}{(1+0)^3}x -  \frac{1}{2!}\frac{3!}{(1+0)^4}x^2 + \frac{1}{3!} \frac{4!}{(1+0)^5}x^3 + ...$$
And simplifying $(1+0) = 1, \frac{(n+1)!}{n!} = (n+1)$ we have
$$ \frac{1}{(1+x)^2} = 1-2x +  3x^2 - 4x^3 + ...$$
As desired
