Shaun,

I am really sorry that I have been too busy lately.The 'gist' of my argument is as follows:

and

The expansion of

Remember this expression.Now the RHS definitely contains 1 (as the first term) and

(as the second term).

It also contains

.

Couple

and

into

The expression I told you to remember also has

and

on the RHS.

Little thought reveals that there is also a term

on the RHS.(which is the product of

and

).This term can also be written as

.This term along with the two mentioned above can be collectively written as

.In a similar manner (using mathematical induction) we can prove that every term of the expansion of

is present in the expansion of

.

Understand carefully what this means. We have not proved that the two expressions are identical.We have merely proved that

where Q is a function of a & b which contains terms not in the expansion of

.We now have to prove that Q=0,that is there is no term in the expansion of

which is not present in the expansion of

, which is where the binomial theorem comes into play.

We seek to prove this by contradiction.Assume Q is not zero.Then Q must be the sum of terms of the form

(as it is a collection of terms in the expansion of

)where p & q are whole numbers.

Consider each such term of Q separately.Put p+q=n

But this term is present in the binomial expansion of

and is hence present in the expansion of

contradicting our defintion of Q according to which Q should not contain any term which is in the expansion for

.