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. 

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).
