$\begin{align} \sum_i (-1)^i \binom{n}{i} D(n,k-i) &= \sum_i (-1)^i \binom{n}{i} \binom{n-1+k-i}{k-i} \\ &= (-1)^k \sum_i \binom{n}{i} \binom{-n}{k-i} \tag{1} \\ &= (-1)^k \binom{0}{k} \tag{2} \end{align}$ which is 1 if $k=0$ and 0 otherwise.
On step (1): I interpret the binomial coefficient $\binom{x}{j}$ as a polynomial in $x$ of degree $j$, via $ \binom{x}{j} = \frac1{j!} \prod_{i=0}^{j-1} (x-i) $ One can check that, under this convention, $\binom{-x}{j} = (-1)^j \binom{x+j-1}{j}$, which gives step (1).
Step (2) uses Vandermonde's convolution, which asserts $ \binom{n+m}{k} = \sum_i \binom{n}{i} \binom{m}{k-i} $ Often this is stated only for nonnegative integers $n$ and $m$; for such values it can be proved combinatorially. To extend it to real $m$, note that under the convention stated above, both sides are polynomials in $m$ of degree $k$, so the fact that they agree at all nonnegative integer values of $m$ implies that they agree for all real $m$. (It would be enough that they agree at $k+1$ distinct values.)
(The best source I know of for techniques like this is Graham, Knuth, and Patashnik's Concrete Mathematics.)