Binomial Coefficient Pattern

Question

Let $n$ $\epsilon$ N and let $k$ $\epsilon$ {0,...,n}. Explain why it follows from ${n \choose k}$ = ${n \choose k-1}$$\frac{n-k+1}{k}$ that ${n \choose k}$ = ($\frac{n}{1}$)($\frac{n-1}{2}$)($\frac{n-2}{3}$)...($\frac{n-k+1}{k}$). We know ${n \choose 0}$=1. Use this to find ${n \choose 1}$. Then use this to find ${n \choose 2}$ and so on and so forth. Notice the pattern.

Answer 1

$${n \choose k}={n \choose k-1}\frac{n-k+1}{k}$$ if you take $k-1$ instead of $k$ in the above identity, you will get $${n \choose k-1}={n \choose k-2}\frac{n-(k-1)+1}{k-1}$$ now apply this into the above one: $${n \choose k}={n \choose k-1}\frac{n-k+1}{k}={n \choose k-2}\frac{n-(k-1)+1}{k-1}\frac{n-k+1}{k}$$ and continue this process until $\binom{n}{k-k}=\binom{n}{0}=1$ is reached.