First we see if it holds for
, therefore true for
Now we use what's called the induction step, we assume that it's true for where k is some natural nubmer.
This gives us
We then prove this for , giving us
Expanding the brackets gives
which can be written as
Therefore it is true for if true for .
You know it is true for 0, therefore true for 1. True for one, therefore true for 2....
Hope this helps