[math]

That just makes the proof short

Because it's not true

I'll prove it for you brainlet out there.

>Assume we are at a Veeky Forums meet up.
>For some reason only people from /pol/ and /tv/ showed up
> in fact there are only m+n people in total at the meetup: m from /pol/ and n from /tv/
> Question : how many sungrounded of i people can we make
> Answer 1: by definition , this is m+n choose i
>Answer 2: on the other hand , any i subgroup will have a certain number of k people from /pol/ and i-k from /tv/. As the k can range from 0 to I we sum and we are done...

[math]
\sum\limits_{i=0}^{k}{{n+i}\choose{i}}={{n+k+1}\choose{k}}
[/math]
Double count this you little shit.

In this case the proof from the book would be a very easy double counting argument for proving Pascal's identity (easy and classic), then trivially applying that by induction.