I found question, that is primary question for my problem. Can't ask my question via comment to the second answer, because have not enough reputation. In proving of $$(1+q)(1+q^2)(1+q^4)\dots(1+q^{{2}^{n}}) = \frac{1-q^{{2}^{n+1}}}{1-q}$$

I got till $1-\left(q^{2^{n+1}}\right)^2$ and thought that it equals to $1-\left(q^{2^{2(n+1)}}\right)=1-\left(q^{4^{n+1}}\right)$, but it's wrong for sure!

Now I want to understand, that I'm right about $(q^{2^{n+1}})^2=(q^{2n}q)^2=q^{2n}q^2=(q^{2^{n+2}})$

If it is possible, please explain me why is it so.

  • 2
    $\begingroup$ $(q^a)^2=q^{2a}$, and let $a=2^{n+1}$. $\endgroup$ – Lord Shark the Unknown May 10 at 17:25
  • $\begingroup$ Yes, you are treating $R=(q^a)^2$ as $W=q^{(a^2)}$ where $a=2^{n+1}.$ $R$ and $W$ are not equal, in general. $\endgroup$ – Thomas Andrews May 10 at 17:27
  • $\begingroup$ Your last line id also confusing. It is not true that $\left(q^{2^{n+1}}\right)^2=(q^{2n}q)^2$ under any interpretation. $\endgroup$ – Thomas Andrews May 10 at 19:27

Letting $m=2^{n+1}$ then you are correct that $m^2=4^{n+1}.$

But the expression is not $q^{(m^2)}$, the expression is $(q^m)^2$, and these two expressions are not equal. In particular, we have that $(q^m)^k=q^{mk}.$

When $k=2$ we have that $(q^m)^2=q^{2m}$ and $2m=2^{n+2}.$

  • $\begingroup$ Just to be sure, that I got it: $q^{4^{n+1}}=q^{2^{n+2}}$? $\endgroup$ – Pavel Stepanov May 11 at 16:08

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