Show that $\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$ I cam across a question in The Book of Proof that states:
Show that $$\binom{n}{3} = \binom{2}{2} + \binom{3}{2} + \binom{4}{2}+...+\binom{n - 1}{2}$$
Which I found the answer to be:
Assume n ≥ 3. Then:
$$\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2} = \binom{n - 2}{3} + \binom{n-2}{2}+ \binom{n - 1}{2} = \binom{2}{2} + \binom{3}{2}+...+\binom{n-1}{2}$$
I have just learned about the binomial theorem, and I am not sure how we got to that answer.
I understand that $\binom{n}{3} = \binom{n - 1}{3} + \binom{n - 1}{2}$ because that is the sum of the two previous rows in the Pascal triangle. But I can't find a simple explanation for the steps that follow or for why we assumed $n \geq 3$

$$S=\sum_{k=2}^{n-1} {k \choose 2} =\text{co-efficient of}~ x^2~ \text{in}\sum_{k=0}^{n-1}(1+x)^k=  \text{co-efficient of} ~x^2~ \text{in} \frac{(1+x)^n-1}{1+x-1}. $$
$$\implies S= \text{co-efficient of} ~x^3~ \text{in} [(1+x)^n-1]={n \choose 3}$$
