Why does Pascal's Triangle give the Binomial Coefficients? An exercise in chapter 2 of Spivak's Calculus (4th ed.) talks about how Pascal's triangle gives the binomial coefficients. It explains this by saying that the relation $\binom{n+1}{k} = \binom{n}{k-1}+\binom{n}{k}$. I'm having trouble seeing how this equation gives rise to Pascal's triangle, so any explanation of what's really going on would be helpful, thanks.

If $(n,k)$ is the $k$th entry of the $n$th row of Pascal's triangle, then we have the following equation from the way Pascal's triangle is built:
$$(n+1,k)=(n,k-1)+(n,k)$$
Notice the similarity with the binomial coefficient identity you mention.  Now, since $(n,1)=(n,n)=1$, and since $\binom{n}{1}=\binom{n}{n}=1$, by induction it follows that $(n,k)=\binom{n}{k}$.
Explicitly, if $(n,k)=\binom{n}{k}$ for all $k=1,\ldots,n$, then we have 
$$(n+1,k)=(n,k-1)+(n,k)=\binom{n}{k-1}+\binom{n}{k}=\binom{n+1}{k}$$
for all $k=1,\ldots,n$, and by definition, we already have $(n+1,n+1)=1=\binom{n+1}{n+1}$
