Proof of the hockey stick/Zhu Shijie identity $\sum\limits_{t=0}^n \binom tk = \binom{n+1}{k+1}$ After reading this question, the most popular answer use the identity
$$\sum_{t=0}^n \binom{t}{k} = \binom{n+1}{k+1},$$
or, what is equivalent,
$$\sum_{t=k}^n \binom{t}{k} = \binom{n+1}{k+1}.$$
What's the name of this identity? Is it the identity of the Pascal's triangle modified.
How can we prove it? I tried by induction, but without success. Can we also prove it algebraically?
Thanks for your help.

EDIT 01 : This identity is known as the hockey-stick identity because, on Pascal's triangle, when the addends represented in the summation and the sum itself are highlighted, a hockey-stick shape is revealed.


First proof
Using stars and bars, the number of ways to put $n$ identical objects to $k$ bins(empty bin allowed) is $\binom{n+k-1}{k-1}$.
If we reduce the number of bins by one, the number of ways to put $n$ identical objects to $k-1$ bins is $\binom{n+k-2}{k-2}$.
We can also count the number of ways to put $n$ identical objects to $k$ bins using the answer for $k-1$ bins. Split them into different cases based on how many objects you put in the first bin.
\begin{array} {|r|r|}\hline \text{Number of objects for first bin} & \text{Number of objects remaining for other bins} & \text{Number of ways to distribute} \\ \hline 0 & n & \binom{n+k-2}{k-2} \\ \hline 1 & n-1 & \binom{n+k-3}{k-2} \\ \hline 2 & n-2 & \binom{n+k-4}{k-2} \\ \hline \vdots & \vdots & \vdots \\ \hline n-2 & 2 & \binom{k}{k-2} \\ \hline n-1 & 1 & \binom{k-1}{k-2} \\ \hline n & 0 & \binom{k-2}{k-2} \\ \hline  \end{array}
Therefore,
$$\binom{n+k-1}{k-1} = \binom{k-2}{k-2} + \binom{k-1}{k-2} + \binom{k}{k-2} +\dots+ \binom{n+k-4}{k-2} + \binom{n+k-3}{k-2} + \binom{n+k-2}{k-2}$$
Rename the variables: Let $m+1 = n+k-1$ and $r+1 = k-1$. We get:
$$\binom{m+1}{r+1} = \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} +\dots+ \binom{m-2}{r} + \binom{m-1}{r} + \binom{m}{r}$$
This is the identity when expanded.
Second proof
Suppose we have $m+1$ people where everyone has different age and we want to select $r+1$ people to be in some committee. There are $$\binom{m+1}{r+1} $$ ways to form the committee.
Alternatively, we can count this by splitting the committee based on who the oldest person in the committee is.
We can have the oldest person to be the oldest in the committee. Then there are $m$ people left and we still need to add $r$ people to be in the committee. So there are $\binom{m}{r}$ ways.
We can have the second oldest person to be the oldest in the committee. Then there are $m-1$ people left since we must exclude the first oldest person and second oldest(already selected) and we still need to add $r$ people to be in the committee. So there are $\binom{m-1}{r}$ ways.
In general, suppose we have the kth oldest person to be the oldest in the committee. Then there are $m-k+1$ possible people left to add(excluding kth oldest) since we must exclude the oldest members up to the $k$th oldest person. We still need to add $r$ people to be in the committee. So there are ${m-k+1 \choose r}$ ways.
In each case, there needs to be at least $r$ people left to choose so the last case is when we select the $(m+1-r)$th oldest person. This will give us ${r \choose r}$ ways.
So
$$\binom{m+1}{r+1} = \binom{r}{r} + \binom{r+1}{r} + \binom{r+2}{r} +\dots+ \binom{m-2}{r} + \binom{m-1}{r} + \binom{m}{r}$$
