Adding and subtracting values that don't exist 
Above is a part of calculation involving the binomial coefficient. I understand the steps. What we have done is that we added and subtracted the same terms (those drawn in a red box) so that nothing changed by this step. But these terms don't exist in the definition of a binomial coefficient.. So is it logical to add or subtract a value that doesn't exist?!  I hope that I were clear in presenting what was unclear to me.
 A: I guess you are used to the definition of the binomial coefficients as
$$\binom{n}{k} = \frac{n!}{k!(n-k)!}$$
However if you cancel out the $(n-k)!$, you get
$$\binom{n}{k} = \frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}$$
And if you write it in this way, you see that there is no problem with having arbitrary values for $n$. For example, you have
$$\binom{\pi}{3} = \frac{\pi(\pi-1)(\pi-2)}{6}$$
Or in your case,
$$\binom{n-1}{n} = \frac{(n-1)(n-2)\cdots ((n-1)-n+1)}{n!}$$
Now notice that the last factor in the numerator is  actually zero; therefore the whole binomial coefficient vanishes.
Another way to get to the same result is to notice that there is a recursion formula,
$$\binom{n}{k+1} = \frac{n-k}{k+1}\binom{n}{k}$$
which is easily checked by inserting the original definition. By assuming this recursion formula to hold further, we then get immediately
$$\binom{n-1}{n} = \binom{n-1}{(n-1)+1} = \frac{(n-1)-(n-1)}{(n-1)+1}\binom{n-1}{n-1} = 0$$
OK, that takes care of the first term, but what about the second? Even with this definition, you'd still get $(-1)!$ in the denominator. There are again at least two ways to tackle this.
The first one is to observe that $\binom{n}{k} = \binom{n}{n-k}$ and just assume that this also holds also outside the original range. Then you get
$$\binom{n-1}{-1} = \binom{n-1}{(n-1)-(-1)} = 0$$
A second way is to solve the recursion formula from above for $\binom{n}{k}$:
$$\binom{n}{k} = \frac{k+1}{n-k}\binom{n}{k+1}$$
Then we get immediately
$$\binom{n-1}{-1} = \frac{-1+1}{n-1-(-1)}\binom{n-1}{-1+1} = 0$$
In any case, you get that those extra terms added are well defined and have the value $0$.
