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.
 A: *

*

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

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:

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

A: These may be more familiar as factorisations of numbers consisting of repetitions of a single digit (a-1) written in base a, by splitting it in half repeatedly. Let's take base 2 (binary) to illustrate.
Let's factorise the number consisting of 16 1's.
$1111111111111111 = 100000001\times 10001 \times 101\times 11$
Starting from the end, each multiplication doubles the number of 1's.
$11\times 101=1111$ is four 1's.
$1111\times 10001=11111111$ is eight 1's.
$11111111\times 100000001=1111111111111111$ is sixteen 1's.
And so on.
In base 10, we get repetitions of the digit 9 instead.
$99\times 101=9999$ is four 9's.
$9999\times 10001=99999999$ is eight 9's.
$99999999\times 100000001=9999999999999999$ is sixteen 9's.
In general, any number consisting of a composite number of repetitions of a single digit can be factorised by splitting it into equal-sized blocks.
A: Let’s start with an example. Let $m=5,n=2$. Now, we will use the formula $$k^2-1=(k+1)(k-1)\tag{1}$$ repeatedly. We notice that $a^{2^m}=a^{2^{m-1}\cdot2}=\left(a^{2^{m-1} }\right)^2$
$$a^{2^m}-1 =\left(a^{2^{m-1} }\right)^2-1=(a^{2^{m-1}}+1) (a^{2^{m-1}}-1)$$ So $$a^{32}-1=(a^{16}+1)(a^{16}-1).$$ Using $(1)$ again, we’ll get $$a^{32}-1=(a^{16}+1)(a^{8}+1)(a^8-1).$$$$= (a^{32}+1)(a^{16}+1)(a^{8}+1)(a^4+1)(a^4-1)$$ etc. Notice that the last two terms can be written as $$(a^{2^2}+1)(a^{2^2}-1)= (a^{2^n}+1)(a^{2^n}-1) $$ for $n=2$.
As an algebraic example, as suggested by user @TomKern, $$ \bbox[5px,border:2px solid red]{(x^{16}-1)=(x^8+1)(x^8-1)=…=(x^8+1)(x^4+1)(x^2+1)(x+1)(x-1)}.$$
Thus, if $m,n\in \mathbb Z_+$, $n<m$, then we can write $$a^{2^m}-1 =\left(a^{2^{m-1} }\right)^2-1=(a^{2^{m-1}}+1) \color{blue}{(a^{2^{m-1}}-1)}$$$$= (a^{2^{m-1}}+1) \color{blue}{(a^{2^{m-2}}+1) (a^{2^{m-2}}-1)}$$$$= (a^{2^{m-1}}+1) (a^{2^{m-2}}+1) \color{red}{(a^{2^{m-3}}+1) (a^{2^{m-3}}-1)}$$ $$=…$$$$= (a^{2^{m-1}}+1) (a^{2^{m-2}}+1) (a^{2^{m-3}}+1)(a^{2^{m-3}}+1)… \color{blue}{(a^{2^{m-k}}-1)(a^{2^{m-k}}+1)} $$ where $k\in${$1,2,3,…,m-1$} so that $m-k=n$ is an integer less than $m$. This train of factors can be stopped anywhere in between, but if continued as long as possible, you’ll finish the sequence with $(a^{2^{m-1}}+1) …(2+1)(2-1)$.


One can also do this the reverse way. For eg, $$(2^{128}+1)(2^{64}+1)(2^{32}+1)…(2+1)$$$$= (2^{128}+1)…(2+1)\cdot1$$$$= (2^{128}+1)…(2+1)(2-1)$$$$= (2^{128}+1)…(2^2+1)(2^2-1)=…=2^{256}-1.$$
A: The idea is a simple tree.
$$x^{2a} - 1$$
splits into (using difference of squares):
$$(x^a + 1)(x^a - 1)$$
And if $a$ is also a power of two (say, $a = 2b$), we can repeat the difference of squares on the 2nd part:
$$(x^{2b} + 1)(x^{2b} - 1)$$
$$(x^{2b} + 1)(x^{b} + 1)(x^{b} - 1)$$
or, in short:
$$x^{4b}-1 = (x^{2b} + 1)(x^{b} + 1)(x^{b} - 1)$$
We can repeat this as many times as we want for every power or two, if $a=c2^k$, each time splitting the difference of squares of the $x^z - 1$ term into $(x^{z/2}+1)(x^{z/2}-1)$, until we can't divide it further:
$$x^{c2^k}-1 = (x^{c2^{k-1}} + 1)(x^{c2^{k-2}} + 1)...(x^{c2^1}+1)(x^{c2^0}+1)(x^{c2^0} - 1)$$
To formally prove this is true, you'd probably end up doing an induction argument.
We then just have to set $c=2^n$ for some $n$, and $m = n+k$, and the result follows.
