What's the Maclaurin Series of $f(x)=\frac{1}{(1-x)^2}$? This function seemed to be pretty much straight forward, but my solution is incorrect.
I have two questions:
1. Where did I make a mistake?
2. I learned that there are shortcuts for finding a series (substitution / multiplication / division / differentiation / integration of both sides). Is there something that I can apply here?
$$f(x)=\frac{1}{(1-x)^2}  \qquad  f(0)=1$$
$$f'(x)=\frac{2}{(1-x)^3}  \qquad  f'(0)=2$$ 
$$f''(x)=\frac{6}{(1-x)^4}  \qquad  f''(0)=6$$ 
$$f'''(x)=\frac{24}{(1-x)^5}  \qquad  f'''(0)=24$$ 
$$f''''(x)=\frac{120}{(1-x)^6}  \qquad  f''''(0)=120$$
The series is then
$$1+\frac{2x}{2!}+\frac{6x^2}{3!}+\frac{24x^3}{4!}+\frac{120x^4}{5!}$$
and when simplified is 
$$1+x+x^2+x^3+x^4$$
But the correct answer is 
$$1+2x+3x^2+4x^3+5x^4 $$

As far as shortcuts go, there are two you might find interesting.
You've probably seen the geometric series several times:
$$ \frac{1}{1-x} = 1 + x + x^2 + x^3 + \ldots $$
Now the first idea is that your function $f(x)$ is the square of this one, so:
$$ f(x) = (1 + x + x^2 + x^3 + \ldots )^2 $$
If the indicated multiplication is carried out, one gets:
$$ f(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots $$
The second idea is to differentiate the geometric series formula:
$$ \frac{d}{dx}(\frac{1}{1-x}) = \frac{1}{(1-x)^2} = f(x) $$
Thus the term-by-term differentiation of the geometric series also gets:
$$ f(x) = 1 + 2x + 3x^2 + 4x^3 + \ldots $$
