Find the coefficient of $x^{100}$ in the power series representing the function $\dots$

Question

Find the coefficient of $x^{100}$ in the power series representing the function: $$f(x)=(x+x^2+x^3+ \cdots) \cdot (x^2+x^3+x^4 \cdots) \cdot (x^3+x^4+x^5 \cdots)$$

Here is what I have so far:
$x+x^2+x^3+ \cdots = x(1+x+x^2+x^3+ \cdots)$ and this equals to the geometric series $\frac{x}{1-x}$
$x^2+x^3+x^4 \cdots = x^2(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^2}{1-x}$
$x^3+x^4+x^5 \cdots = x^3(1+x+x^2+x^3+ \cdots)$ and this equals $\frac{x^3}{1-x}$

So now we have $F(x)=\left(\frac{x}{1-x}\right)\left(\frac{x^2}{1-x}\right)\left(\frac{x^3}{1-x}\right)$
This is the same as $F(x)=x^6(1-x)^{-1}$

Now we can use the following way to expand and find the coefficient in front of $x^{100}$, $(1+x)^{\alpha}=1+\alpha x + {\alpha \choose 2}x^2 + {\alpha \choose 3}x^3 + \dots$

For our function we get $x^6(1+(-1)(-x) + {-1 \choose 2}(-x)^2 + {-1 \choose 3}(-x)^3 + \dots)$, but $-1 \choose \beta$ where $\beta \ge 2$ will always equal $1$ or $-1$ and I do not think I am doing this correct because according to this the coefficient of $x^{100}$ will be either $1$ or $-1$?

What am I doing wrong?

Answer 1

A different conceptual interpretation: We have

$$ F(x) = x^6 (1+x+x^2+\cdots)^3 $$

We can interpret this as a generating function representing a fixed prefix of six balls, plus three (distinguishable) sacks of zero or more balls each. The coefficient of $x^{100}$ is then the number of ways to total $100$ balls. Factoring out the six-ball prefix, that leaves the number of ways to obtain $94$ balls in the three sacks, which we can determine using stars-and-bars as

$$ \binom{96}{2} = 4560 $$

Answer 2

As others have noted in the comments, you should have

$$F(x)=\frac{x^6}{(1-x)^3}=x^6\sum_{n\ge 0}\binom{n+2}2x^n=\sum_{n\ge 6}\binom{n-4}2x^n\;.$$

The general rule here is

$$\frac1{(1-x)^{m+1}}=\sum_{n\ge 0}\binom{m+n}mx^n\;.$$