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.


This is essentially the same as the induction answer already mentioned, but it brings a pictorial perspective so I thought to add it as an alternative answer here.
Here's a restatement of the identity (which you can verify to be equivalent easily): On Pascal's triangle, start from a number (one of the $1$s) on the left edge and move diagonally rightwards and downwards, summing the terms as we go along. We can decide to stop at any point, and the sum of all these terms will be the number directly below and to the left of the final summand.
This actually trivialises the proof of the identity. Note that if we decided to add one more term to the sum (the term to the right of the current sum), on the one hand this "lengthens" the stick by $1$ tile, but on the other hand it adds the term adjacent to the sum---which by definition of Pascal's triangle, produces the number in the tile directly below and horizontally in between the sum and this new term. This can be rigorously translated to the inductive step in a formal induction proof.
 
To illustrate, let's refer to the picture in the question, and focus on the yellow hexagonal tiles. (Note that this is a reflected case of what I described above since it starts from the right edge, but this doesn't affect the discussion.) Currently, we have $1+6+21+56=84$, which is a true identity. If I added $126$ to the LHS, I would also be adding $126$ to the RHS, which by definition gives us $210$, the term below and in between them on Pascal's triangle. Once you really convince yourself of the validity of this argument, the (formal) proof of the identity should come naturally!
