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.


Let me attempt a “story” solution:
Let’s say you are trying to find ways to select $k+1$ integers from the first $n+1$ integers.
We consider all the cases, from where the largest number is $k+1$, up to where the largest number is $n+1$.
It is evident that if the largest number is $k+1$, that there is only one possibility, which is selecting every number from $1$ to $k+1$, and we can write this as $\binom kk$.
More generally, if the largest number is $j\geq k+1$, there are $\binom{j-1}{k}$ selections.
We then sum all of these up, all the way to where the largest number is $k+1$ itself.
