<s>[INST] problem proving: $(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$ I'm trying to prove this, and it is really frustrating, because it seems a really easy problem to prove, however, I'm having a little problem with exponents:
$$(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$$
Hypothesis
$F(x)=(1+q)(1+q^2)(1+q^4)...(1+q^{{2}^{x}}) = \frac{1-q^{{2}^{x+1}}}{1-q}$
The format can be a little problematic here, so, to clarify:
$(1+q^{{2}^{x}})$ = 1+(q^(2^x))
and
$1-q^{{2}^{x+1}}$ = 1 - (q^2^(x+1))
Proof:
$P1 | F(x) = (\frac{1-q^{{2}^{x+1}}}{1-q})(1+q^{{2}^{x+1}}) = \frac{1-q^{{2}^{x+2}}}{1-q} $
$P2| \frac{[(1)-(q^{{2}^{x+1}})][(1)+(q^{{2}^{x+1}})]}{(1-q)} = \frac{1-q^{{2}^{x+2}}}{1-q}$
Here I don't know if I should:
$P3 | \frac{(1-q^{{2}^{x+1}})^2}{1-q} = \frac{1-q^{{2}^{x+2}}}{1-q}| $ applying $ (a+b)(a-b) = a^2 - b^2 $
or
$P3 | \frac{1-q^{{2}^{x+1}+{2}^{x+1}}}{1-q} = \frac{1-q^{{2}^{x+2}}}{1-q} $
This arises because, suposedly, the property goes:
$(a^n)^m = a^{n*m}$
But, it seems, in this problem, the proponderance is different, kinda like this:
$a^{(n^{m})}$
Because, for example, when q=2:
$ F(0) = 1+2^{({2}^{0})} = 1 + 2^{1} = 3 $ 
which seems to be true, since when evaluating RHS:
$ F(0) = \frac{1-2^{{2}^{0+1}}}{1-2} = 3 $ 
 [/INST] Induction on $n$.
For brevity let $\prod_{j=0}^n(1+q^{2^j})=P(q,n)$.
For the case $n=0$ we have $(1-q)P(q,0)=(1-q)(1+q)=1-q^2=1-q^{2^{0+1}}.$
If $(1-q)P(q,n)= 1-q^{2^{n+1}}$ then $$(1-q)P(q,n+1)=(1-q)P(q,n)(1+q^{2^{n+1}})=$$ $$=(1-q^{2^{n+1}})(1+q^{2^{n+1}})=1-(q^{2^{n+1}})^2=1-q^{2^{n+2}}.$$
For example $$(1-q)(1+q)(1+q^2)=(1-q^2)(1+q^2)=1-(q^2)^2=1-q^4.$$
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