How it's possible We can also take advantage of a basic property of exponents: $a^{b*a}$ can be rewritten as $a^{b^c}$. So, we can write $A^N!$ as  $A^{1^{2^3}}$....
$2^{3*2}$ !=  $2^{3^2}$.
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 A: $a^{(b^c)} \ne (a^b)^c$.
$(2^3)^2 = 8^2 = 64$.
$2^{(3^2)} = 2^9 = 256$.
${a^b}^c$ is not a well defined statement as it is not clear if we mean $(a^b)^c = a^{b*c}$ or if we mean $a^{(b^c)}$ which is something else entirely.
As $(a^b)^c $ can be rewritten as $a^{b*c}$ we don't have a need for the concept of $(a^b)^c$ so we usually (but not always) assume ${a^b}^c := a^{(b^c)}$ which does not have a simple rewritten form.  In which case your first line: "$a^{b*c}$ can be rewritten as ${a^b}^c$" is simply not true.  $a^{b*c}$ can be rewritten as $(a^b)^c$ which must certainly is NOT equal to ${a^b}^c= a^{(b^c)}$.
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Following the link, what was written should have been $A^{N!}$ can be rewritten as $(((((A^1)^2)^3)^4.....$ which is true. But it can not be rewritten as $A^{(1^{(2^{(3^{...})})})}$.
In fact $A^{(1^{(2^{(3^{...})})})}= A^{1^K} = A^1 = A$ which is obviously not the intended result.
To be honest, I was too tired to read the entire page so I wasn't sure of their intent.  It will probably work if we interpret the result as meaning.
$2^{24} = 2^{4!} = ((((2^1)^2)^3)^4)$.... which it does.  $((((2^1)^2)^3)^4) = (4^3)^4 = 64^4 = (2^6)^4 = 16777216 =  2^{24}$.
I'm not sure what the point is though.
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Short hand answer:  Exponentiation is not associative.
That's all there is to it. Without associativity expressions are ambiguous.  It's exactly the same issue as the elementary school dilemma of $3 + 4\times 2 = 7\times 2 = 14$ vs.  $3+4\times 2 = 3 + 8 = 11$ but $14 \ne 11$.  Exactly the same issue.
A: Other as for addition and multiplication, in the case of power, parantheses are important.
$$a^{b^c}=a^{(b^c)}\not=(a^b)^c=a^{b\cdot c}.$$
So 
$$x^{n!}=x^{1\cdot2\cdot3\cdot...\cdot n}=(\cdots((x^1)^2)^3\cdots)^n\quad\text{but not}\quad x^{1^{2^{3{\,\dots}}}}$$
A: The author is using a bad notation. What he meant is correctly denoted as
$$a^{bc}=\left(a^b\right)^c$$
hence
$$A^{N!}=\left(\left(\left(A^1\right)^2\right)^3\cdots\right)^N$$
