Sum over k of r-kt choose k by z^k/Proof 2

Proof
From Sum over $k$ of $\dbinom {r - k t} k$ by $\dfrac r {r - k t}$ by $z^k$:


 * $\displaystyle (1): \quad \sum_k \map {A_k} {r, t} z^k = x^r$

where:
 * $\map {A_n} {x, t}$ is the polynomial of degree $n$ defined as:
 * $\map {A_n} {x, t} := \dbinom {x - n t} n \dfrac x {x - n t}$
 * for $x \ne n t$


 * $z = x^{t + 1} - x^t$.

Differentiating $(1)$ $z$:

We have:

Hence: