Help learning and understanding polynomial factorizations A number theory book I'm reading used this factorization as a main step for a proof:

Given $m > n$, integers, $$\left( a^{2^{m}} - 1 \right) = \left( a^{2^{m-1}} + 1 \right)\left( a^{2^{m-2}} + 1 \right) \left( a^{2^{m-3}} + 1 \right) \cdots \left( a^{2^n} + 1 \right) \left( a^{2^n} - 1 \right)$$

Original from the book (in Portuguese):


I'm having a bit of trouble trying to understand how to derive it systematically (in a simple way, without lots of calculations).
If this a known common factorization? Im I missing something?
If so (or not) could one suggest source to learn such factorizations?
Sorry for the possibly relatively low effort question. I just tried to find material on such factorizations and could not find.
Thanks in advance.

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$(x^2 -1) = (x+1)(x-1)$

We know that through experience.  We can explain it through "you see that $+$ in one term and the $-$ in the other... they cause all the middle terms in an expansion to cancel out".   That is $(x+1)(x-1) = x(x-1) + 1(x-1) = x^2 \underbrace{\color{red}{- x + x}}_{\text{cancel out}} -1=x^2 -1$

*

*a) so we we replace $x$ we an power, say $x^k$ then it follows that $(x^{2k}-1)= (x^k +1)(x^k-1)$.  (Because $x^{2k} =(x^k)^2$ and so $((\color{green}{x^k})^2 -1) = (\color{green}{x^k} +1)(\color{green}{x^k}-1)$


*$2^k$ is of course and even number (it is a power of two, after all).  And if $2^k = 2n$ what is $n$?  Well, obviously that means $n =\frac {2^k}2 = 2^{k-1}$.  Increasing an exponent by $1$ will double the value and decreasing the exponent by one will have the number.  This is no surprise.  But what the heck does this have to do with 1. above?


*Since $2^k$ is always even.  $x^{2^k}$ must be a perfect square.  And what is $\sqrt{x^{2^k}}$?  Well it must be $x^{\frac {2^k}2}$.  And what is $\frac {2^k}2 =2^{k-1}$ so $\sqrt{x^{2^k}} = x^{2^{k-1}}$.  This might intuitively not look right because the we are so removed in exponents of exponents but it makes sense.  $\sqrt{x^{2^k}} = x^{\frac {2^k}2}=x^{2^{k-1}}$.  Meanwhile $\sqrt{\sqrt{x^{2^k}}} = \sqrt{x^{2^{k-1}} }= x^{2^{k-2}}$ and so on.


*So $(x^{2^k} -1) = (x^{2^{k-1}} +1)(x^{2^{k-1}} -1)$   and as $x^{2^{k-1}}$ is also a perfect square we get:

$(x^{2^k} -1) = (x^{2^{k-1}} +1)(x^{2^{k-1}} -1)=$
$(x^{2^{k-1}} +1)(x^{2^{k-2} }+ 1)(x^{2^{k-2}}-1 =$
$(x^{2^{k-1}} +1)(x^{2^{k-2}} + 1)(x^{2^{k-3}} + 1)(x^{2^{k-3} } -1)=$
$...$
$(x^{2^{k-1}} +1)(x^{2^{k-2} }+ 1)(x^{2^{k-3}} + 1)(x^{2^{k-4}}+ 1)......(x^{2^{k-m}}+1)(x^{2^{k-m}} -1)=$
$...$
$(x^{2^{k-1}} +1)(x^{2^{k-2}} + 1)(x^{2^{k-3}} + 1)(x^{2^{k-4}}+ 1)......(x^{2^1}+1)(x^{2^{0}} +1)(x^{2^{0}}-1)=$
$(x^{2^{k-1}} +1)(x^{2^{k-2}} + 1)(x^{2^{k-3}} + 1)(x^{2^{k-4}}+ 1)......(x^{2}+1)(x +1)(x-1) $

Your result of $m>n$ that $a^{2^m} -1$ (you had a typo... you typed $a^{2^{m-1}} - 1$ instead) is just a step on the way:

$a^{2^m}-1 = (a^{2^{m-1}} +1)(a^{2^{m-1}} -1)=$
$a^{2^m}-1 = (a^{2^{m-1}} + 1)(a^{2^{m-2}}+1)(a^{2^{m-2}}-1) =$
$a^{2^m}-1 = (a^{2^{m-1}} + 1)(a^{2^{m-2}}+1)....(a^{2^{n+1}}+1)(a^{2^n} +1)(a^{2^n}-1)$.

