Common proof for $(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} $ I'm asking for an alternative (more common?) proof of the following equality, more specifically an alternative proof for the inductive step:
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})=\dfrac{1-x^{2^{n+1}}}{1-x} (x\neq 1)$$
This is how I proved it: 
Basecase: substitute $1$ for $n$, everything works out. 
Inductive step: assume that $$\prod _{i=1}^{n}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x}$$
then $$\prod _{i=1}^{n+1}(1+x^i)=\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1})$$
Let $a$, $b$ and $c$ be positive real numbers, then $\dfrac {c}{a}=b\Leftrightarrow ab=c$, thus 
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=1-x^{2^{n+1}}  \Leftrightarrow \dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$$
$$\dfrac {\left( \dfrac{1-x^{2^{n+2}}}{1-x}\right)}{\left( \dfrac{1-x^{2^{n+1}}}{1-x}\right)}=\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}}$$
Applying polynomial division, we see that indeed $\dfrac {1-x^{2^{n+2}}}{1-x^{2^{n+1}}} = 1+x^{2^{n+1}}$. 
Thus $\dfrac{1-x^{2^{n+1}}}{1-x} (1+x^{n+1}) = \dfrac{1-x^{2^{n+2}}}{1-x}$.
However, the exercise was in a chapter on binomial coefficients and pascal's triangle, furthermore we didn't mention polynomial division in class. Which makes me think that there was another solution that I was "supposed" to see. How was I supposed to prove it?

In the inductive step, you assume that 
$$(1+x)(1+x^2)(1+x^4)...(1+x^{2^n}) = \dfrac{1-x^{2^{n+1}}}{1-x}$$
Which means that
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n}) = 1-x^{2^{n+1}}$$
Multiply by $(1+x^{2^{n+1}})$ and you get
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})(1+x^{2^{n+1}}) = (1-x^{2^{n+1}})(1+x^{2^{n+1}})$$
Using $(a-b)(a+b)=a^2 - b^2$, the left side simplifies to
$$(1-x^{2^{n+1}})(1+x^{2^{n+1}}) = 1 - \left(x^{2^{n+1}}\right)^2 = 1-x^{2^{n+2}}.$$
This means that the previous equation becomes
$$(1-x)(1+x)(1+x^2)(1+x^4)...(1+x^{2^n})(1+x^{2^{n+1}}) = (1-x^{2^{n+2}})$$
and, after dividing by $(1-x)$, you are done. No polynomial division required.
