Question: I am trying to work through the proof I understand it until on this website <http://www.mcs.sdsmt.edu/ecorwin/cs251/bin_thm/bin_thm.html> :

Pascal's identity does not apply for $k = 0$ or for $k = n$. In our sum, this means we need to split out the $j = 0$ and $j = k + 1$ terms before applying Pascal's identity.

Why does k=o and k=n correspond to $j=0$ and $j=k+1$?

Now we have the problem of matching the two sums from the left side with the two from the right side. First, notice that the out front on the right side can be thought of as the $j = 0$ term in the first sum. Next, the y $k+1$ out front on the right side can be thought of as the $j = k + 1$ term in the second sum. 

Why can you think of these terms equaling $j=0$ in the first sum, why not for the second sum? And the same for $j=k+1$?

Answer: 
There are several proofs on the wikipedia page on the binomial theorem.

Later addition: Regarding the proof you cited, Pascal's identity also holds for those endpoints $k=0$ and $k=n$ (all you have to do is define the binomial to be $0$ when $k>n$ or $k<0$).
Pascals Identity is: $\dbinom {k+1} j = \dbinom k{j-1}+\dbinom k j$

Now, if it applied at $j=0$ or $j=k+1$ we would need an evaluation for: $$\dbinom{k}{-1}=\dbinom k{k+1}=\dfrac{k!}{(-1)!~(k+1)!}$$

Okay, *by convention* this is *usually* taken as zero.   But *by definition* it includes a factor of the factorial of minus one, which is indefinite.   So they just... sidestepped the issue and explicitly evaluated the end terms, since *those* are uncontroversial.

$$\begin{align}\sum\_{j=0}^{k+1} \binom{k+1}{j} x^{k+1-j}y^j ~ & = ~ \binom{k+1}0 x^{k+1}+\binom{k+1}{k+1} y^{k+1}+\sum\_{j=1}^k \binom {k+1}j x^{k+1-j}y^j 
\\ & = ~ x^{k+1}+y^{k+1}+\sum\_{j=1}^k\left[\binom{k}{j-1}+\binom{k}{j}\right]x^{k+1-j}y^j
\\ & \quad\vdots\quad \text{et cetera}
\end{align}$$