Let's give a combinatorial proof.
$\binom{n+1}{k+1}$ is the number of subsets of $[n] = \{0,1,2, \ldots, n\}$ having exactly $k+1$ elements. Looking at the smallest or largest element of such a subset makes sense.
Let's look at the second smallest element instead. It can be any number from $1$ (for subsets which have $0$ and $1$ as their two smallest elements) to $n-k+1$ (for subsets which have $n-k+1, \ldots, n-1, n$ as their $k$ largest elements).
So $$\binom{n+1}{k+1} = \sum_{i=1}^{n-k+1} \textrm{number of subsets having $i$ as 2nd smallest element}.$$
So how many $(k+1)$-element subsets of $[n]$ have $i$ as 2nd smallest elements?
Well, the recipe to make such a subset is clear:
- Chose any number in $0, 1, \ldots, i-1$ to serve as smallest element.
- Pick $i$ as second smallest element.
- Pick any $(k-1)$-element subset of $\{i+1, \ldots, n\}$ to serve as the $n-1$ elements larger than $i$.
In the first step, you have $i$ choices, the second step involves no choice and you have $\binom{n-i}{k-1}$ choices in the last step, because $\{i+1, \ldots, n\}$ has $n-i$ elements.
So the number of $(k+1)$-element subsets of $[n]$ having $i$ in second position is $i \binom{n-i}{k-1}$, which proves that
$$\binom{n+1}{k+1} = \sum_{i=1}^{n-k+1} i \binom{n-i}{k-1}.$$
A slight generalisation of this proof shows that if $a+b=k$
$$ \binom{n+1}{k+1} = \sum_i \binom{i}{a}\binom{n-i}{b}:$$
you only have to look at the $(a+1)$-th largest element of the $(k+1)$-element subsets of $n+1$. We have just done the $a=1$ case.
Note that $\binom{n+r}{r}=\binom{n+r}{n}$ is the number of subsets of $\{1,2,\ldots,n+r\}$ of size $n$. On the other hand, for $i=0,1,2,\ldots,r$, $\binom{n+i-1}{i}=\binom{n+i-1}{n-1}$ is the number of subsets of $\{1,2,\ldots,n+r\}$ of size $n$ whose largest element is $n+i$.
Best Answer
Imagine the first $n + 1$ numbers, written in order on a piece of paper. The right hand side asks in how many ways you can pick $k+1$ of them. In how many ways can you do this?
You first pick a highest number, which you circle. Call it $s$. Next, you still have to pick $k$ numbers, each less than $s$, and there are $\binom{s - 1}{k}$ ways to do this.
Since $s$ is ranging from $1$ to $n+1$, $t:= s-1$ is ranging from $0$ to $n$ as desired.