Recall that if |x| < 1, then
[tex]\displaystyle \frac1{1-x} = \sum_{k=0}^\infty x^k[/tex]
So if |a + bxᵐ| < 1, then
[tex]\displaystyle \frac{a+bx^m}{1 - (a+bx^m)} = \sum_{q=1}^\infty (a+bx^m)^q = \sum_{q=1}^\infty a^q \left(1 + \frac{bx^m}a\right)^q[/tex]
By the binomial theorem,
[tex]\displaystyle \left(1 + \frac{bx^m}a\right)^q = \sum_{r=0}^q \binom qr \left(\frac{bx^m}a\right)^r[/tex]
The binomial coefficient is defined as
[tex]\dbinom nk = \dfrac{n!}{k!(n-k)!}[/tex]
if 0 ≤ k ≤ n, and 0 otherwise. Then we can rewrite the sum as
[tex]\displaystyle \left(1 + \frac{bx^m}a\right)^q = \sum_{r=0}^\infty \binom qr \left(\frac{bx^m}a\right)^r[/tex]
and pulling out the first term,
[tex]\displaystyle \left(1 + \frac{bx^m}a\right)^q = 1 + \sum_{r=1}^\infty \binom qr \left(\frac{bx^m}a\right)^r[/tex]
Finally,
[tex]\dbinom qr = \dfrac{q!}{r! (q-r)!} \\\\\\ = \dfrac{q(q-1)(q-2)\cdots(q-(r-1))(q-r)(q-(r+1))\cdots3\cdot2\cdot1}{r! (q-r) (q-(r+1)) (q-(r+2))\cdots3\cdot2\cdot1} \\\\\\ = \dfrac{q(q-1)(q-2)\cdots(q-(r-1))}{r!} \\\\ = \dfrac{\prod\limits_{i=0}^{r-1}(q-i)}{\Gamma(r+1)}[/tex]
since n! = Γ(n + 1). (I think the given upper limit in the product may be a mistake.)
