How do I change my result into the pascal identity? see details please. Use the defintion of binomial theorem to prove the identy.
$$\binom{n}{k} = \binom{n-1}{k}+\binom{n-1}{k-1}$$
The definition of the binomial theorem
$$\binom{n}{k} = \frac{\prod_{i=0}^{k-1}(n-i)}{k!}$$
$$\binom{n-1}{k} = \frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}$$
$$\binom{n-1}{k-1} = \frac{\prod_{i=0}^{k-2}(n-1-i)}{k-1!}$$
$$\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{\prod_{i=0}^{k-2}(n-1-i)}{k-1!}$$
Then I come to the result 
$$\frac{\prod_{i=0}^{k-2}(n-1-i)(n)}{k-2!(k-1)k}$$
$$\frac{\prod_{i=0}^{k-2}(n-1-i)}{k!} \cdot n$$
What can I do to make this result into 
$$\frac{\prod_{i=0}^{k-1}(n-i)}{k!}$$
 A: Open directly the right hand side:
$$\binom{n-1}k+\binom{n-1}{k-1}=\frac{(n-1)!}{k!(n-k-1)!}+\frac{(n-1)!}{(k-1)!(n-k)!}=$$$${}$$
$$=\frac{(n-1)!}{(k-1)!(n-k-1)!}\left[\frac1k+\frac1{n-k}\right]=\frac{{(n-1)!}}{\color{red}{(k-1)!}\color{green}{(n-k-1)!}}\cdot\frac n{\color{red}k\color{green}{(n-k)}}=$$$${}$$
$$=\frac{n!}{k!(n-k)!}=\binom nk$$
A: I think the DonAntonio's idea is the best way to approach the problem, but if you really want finish your idea then take a look. You missed the index at the second term (in red).
$$\binom{n-1}{k} = \frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}$$
$$\binom{n-1}{k-1} = \frac{\prod_{i=0}^{\color{red}{k-2}}(n-1-i)}{(k-1)!}$$
So,
$$\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{\prod_{i=0}^{k-2}(n-1-i)}{(k-1)!}=\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{k\cdot\prod_{i=0}^{k-2}(n-1-i)}{k!}=\\
=(n-1-(k-1)+k)\cdot \frac{\prod_{i=0}^{k-2}(n-1-i)}{k!}=n\cdot\frac{\prod_{i=0}^{k-2}(n-1-i)}{k!}=\frac{\prod_{i=0}^{k-1}(n-i)}{k!}={n \choose k}$$
A: $\binom {n}{k}$ is the co-efficient of $x^k$ in $(1+x)^n$, which is equal to the co-efficient of $x^k$ in $$(1+x)^{n-1}+(1+x)^{n-1}x$$ which is the sum of the co-efficient of $x^k$ in $(1+x)^{n-1}$ and the co-efficient of $x^{k-1}$ in $(1+x)^{n-1},$ which is $\binom {n-1}{k}+\binom {n-1}{k-1}.$
A: it cleat that $$\frac{\prod_{i=0}^{k-1}(n-1-i)}{k!}+\frac{\prod_{i=0}^{k-2}(n-1-i)}{k-1!}=\frac{n\prod_{i=0}^{k-2}(n-1-i)}{k!}$$
put (j=i+1) we obtain
$$n\frac{\prod_{j=1}^{k-1}(n-j)}{k!}=\frac{\prod_{j=0}^{k-1}(n-j)}{k!}$$
which the result.
