I found the following formula at the end of the first chapter of Probability: For the Enthusiastic Beginner by David Morin:
$$\binom{n}{k} = \binom{n-1}{k-1} + \binom{n-2}{k-1} + \binom{n-3}{k-1} + \cdots + \binom{k+1}{k-1} + \binom{k}{k-1} + \binom{k-1}{k-1}$$
As an exercise I decided to prove it.
Here is what I did:
First I wrote down the definition of Pascal's identity:
$$\binom{n-1}{k-1} + \binom{n-1}{k} = \binom{n}{k} \implies \binom{n-1}{k-1} = \binom{n}{k} - \binom{n-1}{k}$$
Looking at the above-mentioned definition we can see that in fact by rewriting Pascal's identity successively and by decreasing each time the value of $n$ by $1$ until $n$ becomes $k-1$ we can build at the end the very same formula we want to prove, that is,
\begin{alignat*}{4} & \text{Pascal's identity} &&\implies \binom{n-1}{k-1} &&= \binom{n}{k} &&- \binom{n-1}{k} \\[5pt] & \text{Reduce $n$ by $1$ in the above line} &&\implies \binom{n-2}{k-1} &&= \binom{n-1}{k} &&- \binom{n-2}{k} \\[5pt] & \text{Reduce $n$ by $1$ in the above line} &&\implies \binom{n-3}{k-1} &&= \binom{n-2}{k} &&- \binom{n-3}{k} \\[5pt] & \text{Reduce $n$ by $1$ in the above line} &&\implies \binom{n-4}{k-1} &&= \binom{n-3}{k} &&- \binom{n-4}{k} \\[5pt] & && \vdots && \\[5pt] & \text{Reduce $n$ by $1$ in the above line} &&\implies \binom{k+2}{k-1} &&= \binom{k+3}{k} &&- \binom{k+2}{k} \\[5pt] & \text{Reduce $n$ by $1$ in the above line} &&\implies \binom{k+1}{k-1} &&= \binom{k+2}{k} &&- \binom{k+1}{k} \\[5pt] & \text{Reduce $n$ by $1$ in the above line} &&\implies \binom{k}{k-1} &&= \binom{k+1}{k} &&- \binom{k+1}{k} \\[5pt] & \text{Reduce $n$ by $1$ in the above line} &&\implies \binom{k-1}{k-1} &&= \binom{k}{k} &&- \binom{k-1}{k} \end{alignat*} Now if we add the left side and right side of the above identities we obtain: $$\sum_{i=k-1}^{n-1}\binom{i}{k-1} = \binom{n}{k} - \binom{k-1}{k}$$ Given that by definition $\binom{k-1}{k} = 0$ we can say: $$\sum_{i=k-1}^{n-1}\binom{i}{k-1} = \binom{n}{k}$$
Which is what we wanted to prove.
Is my solution correct?