How are the factors obtained at *this point* in the solution of proving $2^{2^n}+ 1 \mid 2^{2^m} -1$ I understood everything up until "BUT".  My question is how are the factors on the right side for the bolded line of equation under the word "BUT" obtained?  I tried to get these factors using property (5), but I was unsuccessful, can someone please show me how to get those factors?  Thanks in advance.
This is example 2 from "Problems of Number Theory in Mathematical Competitions" by You Hong-Bing on page 2.
Some properties the example will be using are: 
(1) If $b\mid c$, and $c\mid a$, then $b\mid a$, that is, divisibility is transitive.
(5) If $n$ is a positive odd number, then
$x^n-y^n= (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$
The problem, example 2 is as follows: 
Let $\mathbf{m > n}$ and $\mathbf{n\ge 0}$, show that $\mathbf{(2^{2^n}+1)\mid(2^{2^m}-1)}$ 
The following is the proof from the book: 
Take $x=2^{2^{n+1}}$,$\,\,y=1$ in factorization (5), and substitute $n$ by $2^{m-n-1}$
WE GET: 
$2^{2^m}-1 = (2^{2^{n+1}}-1) \left[(2^{2^{n+1}})^{2^{m-n-1}-1} +\ldots +  2^{2^{n+1}}+1 \right]$
THUS,
$(2^{2^{n+1}}-1) \mid (2^{2^m}-1)$
BUT,
$\mathbf{(2^{2^{n+1}}-1) = (2^{2^n} - 1)(2^{2^n} + 1)}$
This is the equation I'm having trouble with so everything else after this didn't make much sense to me
HENCE,
$(2^{2^n} + 1)\mid(2^{2^{n+1}} - 1)$
FURTHER, BY PROPERTY (1) WE HAVE: 
$(2^{2^n} +1)\mid(2^{2^m} - 1)$
REMARK: sometimes it is difficult to prove $b\mid a$ directly when dealing with divisibility problems.  Therefore, we can attempt to choose an "intermediate number" $c$ and prove $b\mid c$ and $c\mid a$ first, then use property (1) of divisibility to deduce the conclusion.

More generally,
$x^{2^m}-1
=(x-1)\prod_{n=0}^{m-1}(x^{2^n}+1)
$.
For this,
put $x = 2$.
Proof.
For $m=1$ this is
$x^{2^1}-1
=(x-1)\prod_{n=0}^{0}(x^{2^n}+1)
$
or
$x^2-1
=(x-1)(x+1)
$.
If it is true for $m$ then
$\begin{array}\\
(x-1)\prod_{n=0}^{(m+1)-1}(x^{2^n}+1)
&=(x-1)\prod_{n=0}^{m}(x^{2^n}+1)\\
&=(x-1)(x^{2^m}+1)\prod_{n=0}^{m-1}(x^{2^n}+1)\\
&=(x^{2^m}+1)(x-1)\prod_{n=0}^{m-1}(x^{2^n}+1)\\
&=(x^{2^m}+1)(x^{2^m}-1)
\qquad\text{(induction hypothesis)}\\
&=x^{2^{m+1}}-1
\qquad (x^a+1)(x^a-1)=x^{2a}-1
\text{ and } 2(2^m) = 2^{m+1}\\
\end{array}
$
