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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 $

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Define $f(q)=(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}})$,

We calculate

\begin{align*} (1-q)f(q)&= (1-q)(1+q)(1+q^2)(1+q⁴)...(1+q^{{2}^{n}}) \\ &=(1-q^2)(1+q^2)(1+q^{2^2})...(1+q^{{2}^{n}})\\ &=(1-q^{2^2})(1+q^{2^2})...(1+q^{{2}^{n}})\\ &\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \vdots \\ &=1-q^{2^{n+1}} \end{align*} Now divide by $1-q$ and you get the identity.

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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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