Prove for integer $k \ge 0$, $\frac1{(1 - x)^{k+1}}= \sum_{i=k}^{\infty}\binom{i}{k}x^{i-k}$ Prove for integer $k \ge 0$, $$\frac1{(1 - x)^{k+1}}= \sum_{i=k}^{\infty}\binom{i}{k}x^{i-k}$$
I've looked up resources for generating functions. However, this specific example seems to be difficult to prove.
 A: You can prove this directly! Begin with the (formal) geometric series
$$
\frac{1}{1-x} = \sum_{i=0}^{\infty} x^i.
$$
Note that
$$
\frac{d}{dx} \left[ \frac{1}{1-x} \right]= \frac{1}{(1-x)^2}, 
$$
and by induction,
$$
\frac{d^k}{dx^k} \left[ \frac{1}{1-x} \right] = \frac{k!}{(1-x)^{k+1}}, 
$$
for any $k \geq 0$.
So it suffices to calculate the $k$th derivative of the geometric series. Note that for all orders $0 \leq i < k$,
$$
\frac{d^k}{dx^k} \left[ x^i \right] = 0, 
$$
and for $i \geq k$
$$
\frac{d^k}{dx^k} \left[ x^i \right] 
= \frac{k \cdot (k-1) \cdots (k-i+1)}{i \cdot (i-1) \cdots 1} \, x^{i-k}
= \frac{k!}{(i-k)!} \, x^{i-k}.
$$
Thus,
\begin{align}
\frac{1}{(1-x)^{k+1}} 
&= \frac{1}{k!} \frac{d^k}{dx^k} \left[ \frac{1}{1-x} \right] \\[2pt]
&= \frac{1}{k!} \frac{d^k}{dx^k} \left[ \sum_{i=0}^{\infty} x^i \right] \\[2pt]
&= \frac{1}{k!} \frac{d^k}{dx^k} \left[ \sum_{i=k}^{\infty} x^i \right] \\[2pt]
&= \sum_{i=k}^{\infty} \frac{1}{k!} \frac{d^k}{dx^k} \left[ x^i \right] \\[2pt]
&= \sum_{i=k}^{\infty} \frac{i!}{k!\,(i-k)!} \, x^{i-k} \\[2pt]
&= \sum_{i=k}^{\infty} \binom{i}{k} \, x^{i-k}, 
\end{align}
as desired.
