As part of a proof that $$(1-x)^n\cdot \left ( \frac{1}{1-x} \right )^n=1$$ in the context of generating functions it states that $$\sum_{i=0}^k(-1)^i\binom{n}{i}D(n,k-i))=0$$ where $D(n,k)$ is defined as $$D(n,k)=\binom{n-1+k}{k}=\binom{n-1+k}{n-1}\;.$$
I don't understand the above step, which is the last in the proof.
$$(1-x)^n=\sum_{i=0}^{\infty}(-1)^i\binom{n}{i}x^i$$
and
$$\left ( \frac{1}{1-x} \right )^n=\sum_{i=0}^{\infty}D(n,i)x^i$$