Pascal's formula and Binomial Coefficients

Prove Pascal's formula by substituting the values of the binomials coefficient as given in the equation $$\binom nk= \frac{n!}{k!\,(n-k)!} = \frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots1}.$$

I guess I just don't know what values I'm supposed to be substituting in.

$$ {n\choose k}=\frac{n(n-1)\dots(n-k+1)}{k!}=\frac{(n-1)\dots(n-k+1)}{(k-1)!}\frac{n}{k}\\ =(\frac{k}{k}+\frac{n-k}{k})\frac{(n-1)\dots(n-k+1)}{(k-1)!}\\ =\frac{(n-1)\dots(n-k+1)}{(k-1)!}+\frac{(n-1)\dots(n-k+1)(n-k)}{k!}\\ ={n-1\choose k-1}+{n-1\choose k} $$

$$\binom nk=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!(n-(n-k))!}=\binom{n}{n-k}....[1]$$ $$\binom nk=\frac{n(n-1)!}{k(k-1)!(n-k)!}=\frac{ n}{k}\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}=\frac{n}{k}\binom {n-1}{k-1}...[2]$$ from [1] and [2] we have $$\binom {n}{k}=\binom {n}{n-k}=\frac {n}{n-k}\binom {n-1}{n-k-1}=\frac{n-k+k}{n-k} \binom {n-1}{(n-1)-(n-k-1)}=$$ $$=\left(1+\frac{ k}{n-k}\right) \binom {n-1}{k}=\binom {n-1}{k}+\frac{ k}{n-k}\binom {n-1}{k}=$$ $$=\binom {n-1}{k}+\frac{ k}{n-k}\frac{n-1}{k}\binom {n-2}{k-1}=\binom {n-1}{k}+\frac{n-1}{n-k}\binom {n-2}{(n-2)-(k-1)}=$$ $$=\binom {n-1}{k}+\frac{n-1}{n-k}\binom {n-2}{n-k-1}=\binom {n-1}{k}+\binom {n-1}{n-k}=$$ $$=\binom {n-1}{k}+\binom {n-1}{(n-1)-(n-k)}=\binom {n-1}{k}+\binom {n-1}{k-1}$$