Prove Pascal's Rule Algebraically I am trying to prove Pascal's Rule algebraically but I'm stuck on simplifying the numerator. This is the last step that I have, but I'm not sure where to go from here
$$=\frac{\left[(k-1)(n-k)!+k(n-1-k)!\right]\times(n-1)!}{k!(n-k)!}$$

I will assume that you are trying to show that $\binom{n}{k}=\binom{n-1}{k}+\binom{n-1}{k-1}$, where $n$ and $k$ are each $\ge 1$. Expressing the right-hand side in terms of factorials, we get
$$\frac{(n-1)!}{k!(n-k-1)!}+\frac{(n-1)!}{(k-1)!(n-k)!}.$$
We want to bring the above expression to a common denominator. So we multiply top and bottom of the first term by $n-k$, and top and bottom of the second term by $k$. We get
$$\frac{(n-1)!(n-k)+(n-1)!k}{k!(n-k)!}.$$
The bottom now looks nice. The two terms on top have a common factor of $(n-1)!$. So the top can be rewritten as
$$(n-1)![(n-k)+k)],$$ 
which is $(n-1)!n$, that is, $n!$. Thus we end up with $\frac{n!}{k!(n-k)!}$, which is $\binom{n}{k}$. 
