The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne 1$, equals The expression
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$
where $q\ne 1$, equals
(A) $\frac{1-q^{128}}{1-q}$
(B) $\frac{1-q^{64}}{1-q}$     
(C) $\frac{1-q^{2^{1+2+\dots +6}}}{1-q}$     
(D) none of the foregoing expressions
What I have done
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ is a polynomial of degree $127$. Now the highest degree of the polynomial in option (A), (B) and (C) is $127$, $63$ and $41$ respectively. And therefore (A) is the correct answer.
I get to the correct answer but I don't think that my way of doing is correct. I mean what if, if option  (B) was $\frac{1+q^{128}}{1-q}$.
Please show how should I approach to the problem to get to the correct answer without any confusion. 
 A: $$(1+q)(1+q^2)=1+q+q^2+q^3=\frac{1-q^4}{1-q}$$
So,
$$(1+q)(1+q^2)(1+q^4)=\frac{(1-q^4)(1+q^4)}{1-q}=\frac{1-q^8}{1-q}$$
So,
$$(1+q)(1+q^2)(1+q^4)(1+q^8)=\frac{(1-q^8)(1+q^8)}{1-q}=\frac{1-q^{16}}{1-q}$$
Can you see the pattern?
A: HINT: Multiply $1-q$ by $1+q$, you obtain $1-q^2$, which multiplied by $1+q^2$ gives $1-q^4$ and so on.
(And: yes, (A) is the correct answer).
A: Let $$P = (1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64}).$$
One has $$(1-q)P = (1-q)(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})
= 1 - q^{128}.
$$
So, one gets $$P = \frac{1-q^{128}}{1-q}$$ 
A: Consider the numerator $1-q^{128}$ as a difference of two squares,  so it is equal to $(1+q^{64})(1-q^{64})$  Now the second term is the difference of two squares, so you can proceed:  $(1+q^{64})(1-q^{64})=(1+q^{64})(1+q^{32})(1-q^{32})$.
Continuing downwards as more and more difference of squares, you end up getting $(1-q^{64})=(1+q^{64})(1+q^{32})(1+q^{16})(1+q^{8})(1+q^{4})(1+q^{2})(1+q^{1})(1-q^{1})$$
As long as $q\ne 1$,  $q-1\ne =0$, so we can divide and get your result.
A: $a^n – b^n = (a – b)(a^{n–1} +a^{n–2}b + a^{n–3}b^2 + ··· + ab^{n–2} + b^{n–1})$
If $n$ is odd,$a^n+b^n = (a+b)(a^{n–1}-a^{n–2}b + a^{n–3}b^2 - ··· - ab^{n–2} + b^{n–1})$
