Properties of a sequence of sums of binomials I have encountered the following sequence of alternating sums of binomials and I am wondering whether there is a nicer way to write every element and/or are there some nice properties about it.
So, if you have any idea about it, please let me know.
First of all pick two positive natural numbers $n, r\in\mathbb{N}, 0<r<n$.
Then, for all $j$, $0\leq j\leq r$, define 
$$a_j:=\sum_{i=0}^j(-1)^{i}\binom{j}{i}\binom{n-i-1}{r-i-1}$$
Note that, using the standard convention $0!=1$, the previous formula is defined also for $j=0$.
The sequence I am interested in is the sequence of these $(a_j)$'s, varying $n, r$ and obviously $j$.
In particular at some point I will sum over $j$.
So the first terms of this sequence are
$$\begin{aligned}
a_0&=\binom{n-1}{r-1}\\
a_1&=\binom{n-1}{r-1}-\binom{n-2}{r-2}\\
a_2&=\binom{n-1}{r-1}-2\binom{n-2}{r-2}+\binom{n-3}{r-3}\\
a_3&=\binom{n-1}{r-1}-3\binom{n-2}{r-2}+3\binom{n-3}{r-3}-\binom{n-4}{r-4}\\
a_4&=\binom{n-1}{r-1}-4\binom{n-2}{r-2}+6\binom{n-3}{r-3}-4\binom{n-4}{r-4}+\binom{n-5}{r-5}\\
\ldots
\end{aligned}$$
As you can see, it is very elegant, so I think that someone already studied it, and probably it is related to some property of the Pascal's triangle.
 A: Here is a simplification: The following formula is valid.

$$\sum_{i=0}^{j}(-1)^i\binom{j}{i}\binom{n-i-1}{r-i-1}=\binom{n-j-1}{r-1}\qquad\qquad0<r<n,\quad0\leq j\leq r$$

Proof by induction: Let $j=0$ then $LHS=RHS=\binom{n-1}{r-1}$. Now assume the formula is valid for $j=k$.
Induction Step: $j \rightarrow k+1$:
\begin{align}
\sum_{i=0}^{k+1}&(-1)^i\binom{k+1}{i}\binom{n-i-1}{r-i-1}\\
&=\binom{n-1}{r-1}+\sum_{i=1}^{k}(-1)^i\binom{k+1}{i}\binom{n-i-1}{r-i-1}+(-1)^{k+1}\binom{n-k-2}{r-k-2}\\
&=\binom{n-1}{r-1}+\sum_{i=1}^{k}(-1)^i\left(\binom{k}{i}+\binom{k}{i-1}\right)\binom{n-i-1}{r-i-1}\\
&\qquad+(-1)^{k+1}\binom{n-k-2}{r-k-2}\\
&=\binom{n-1}{r-1}+\left(\binom{n-k-1}{r-1}-\binom{n-1}{r-1}\right)+\sum_{i=1}^{k}(-1)^i\binom{k}{i-1}\binom{n-i-1}{r-i-1}\\
&\qquad+(-1)^{k+1}\binom{n-k-2}{r-k-2}\qquad\qquad(\star)\\
&=\binom{n-k-1}{r-1}+\sum_{i=0}^{k}(-1)^{i+1}\binom{k}{i}\binom{n-i-2}{r-i-2}\\
&=\binom{n-k-1}{r-1}-\binom{n-k-2}{r-2}\qquad\qquad(\star)\\
&=\binom{n-(k+1)-1}{r-1}
\end{align}
with induction hypothesis used in $(\star)$
