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