How do I use the Binomial expansion with a negative expression that is not a real number

Question

How would I go above proving that

$\sum_{x=k}^\infty{x-1 \choose k-1}(qe^t)^{x-k}=(1-qe^t)^{-k}$

I can't seem to get the Binomial theorem in the correct format to easily see that these are equivalent

Answer 1

Just use the fact that $\binom{-n}{k} = (-1)^k \binom{n + 1 - k}{n - 1}$:

$\begin{align*} \sum_{x \ge k} \binom{x - 1}{k - 1} (q e^t)^{x - k} &= \sum_{j \ge 0} \binom{j + k - 1}{k - 1} (q e^t)^j \\ &= \sum_{j \ge 0} (-1)^j \binom{-k}{j} (q e^t)^j \\ &= (1 - q e^t)^{-k} \end{align*}$

To prove the identity used:

$\begin{align*} \binom{-n}{k} &= \frac{(-n)^{\underline{k}}}{k!} \\ &= \frac{(-1)^k n^{\overline{k}}}{k!} \\ &= (-1)^k \frac{(n + k - 1)^{\underline{k}}}{k!} \\ &= (-1)^k \binom{n + k - 1}{k} \\ &= (-1)^k \binom{n + k - 1}{n - 1} \end{align*}$

Here we use falling and rising factorial powers:

$\begin{align*} n^{\underline{k}} &= n (n - 1) \dotsm (n - k + 1) \\ n^{\overline{k}} &= n (n + 1) \dotsm (n + k - 1) \end{align*}$