$(q^{2^{n+1}})^2$ question to understand another question

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.

• $(q^a)^2=q^{2a}$, and let $a=2^{n+1}$. – Lord Shark the Unknown May 10 at 17:25
• 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. – Thomas Andrews May 10 at 17:27
• 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. – 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 $$2m=2^{n+2}.$$
• Just to be sure, that I got it: $q^{4^{n+1}}=q^{2^{n+2}}$? – Pavel Stepanov May 11 at 16:08