How are "(2)" , "(3)" and "(4)" of the "hypergeometric distribution" combined to create the "(5)" or binomial distribution for large N? Following up to Asymptotic behavior of combinations: approximating Hypergeometric by Binomial.

I'm going to use $k$ instead of $x$, for notational preferences on my part.
  For $k\leq n$,
  $$\begin{align}
\mathbb{P}\{X=k\}&=\frac{\binom{Np}{k}\binom{Nq}{n-k}}{\binom{N}{n}}
\\ &= \frac{(Np)!}{{\color{red}{k!}}(Np-k)!}\cdot\frac{(Nq)!}{{\color{red}{(n-k)!}}(Nq-n+k)!}\cdot\frac{{\color{red}{n!}}(N-n)!}{N!}\\
&= {\color{red}{ \binom{n}{k} }}\cdot \frac{(Np)!}{(Np-k)!}\cdot\frac{(Nq)!}{(Nq-n+k)!}\cdot\frac{(N-n)!}{N!}\tag{1}
\\ &={n \choose k} A*B*C\end{align}$$
Now, as $N\to \infty$ and all other parameters are fixed, we have
  $$
\frac{1}{(Np)^{k}}\frac{(Np)!}{(Np-k)!} = \frac{(Np)(Np-1)\cdots(Np-k+1)}{(Np)^{k}} \xrightarrow[N\to\infty]{} 1\tag{2}
$$
  while 
  $$
\frac{1}{(Nq)^{n-k}}\frac{(Nq)!}{(Np-n+k)!} = \frac{(Nq)(Nq-1)\cdots(Nq-n+k+1)}{(Nq)^{n-k}} \xrightarrow[N\to\infty]{} 1\tag{3}
$$
  and
  $$
N^n\frac{(N-n)!}{N!}= \frac{N^n}{N(N-1)\cdots(N-n+1)} \xrightarrow[N\to\infty]{} 1\tag{4}
$$
  so that, combining them all,
  $$
\mathbb{P}\{X=k\} \operatorname*{\sim}_{N\to\infty} \binom{n}{k} \frac{(Np)^k(Nq)^{n-k}}{N^n} =\boxed{\binom{n}{k} p^k q^{n-k}} \tag{5}
$$
  where $\operatorname*{\sim}_{N\to\infty}$ denotes the standard asymptotic equivalence (Landau notation). 

In "(2)" why are we multiplying by $$\frac{1}{(Np)^k}?$$
In "(3)" why are we multiplying by $$\frac{1}{(Nq)^{n-k}}?$$.
In "(4)" why are we multiplying by $$N^n?$$
Why for "(5)" throw out $A$ and replacing $A$ with the reciprocal of $$\frac{1}{(Np)^{k}}?$$
Why for "(5)" throw out $B$ and replacing that with the reciprocal of $$\frac{1}{(Nq)^{n-k}}?$$
Why for "(5)" throwing out C and replacing C with the reciprocal of $$N^n?$$
 A: Your first three questions are related to what can be done to get limits of $1$ rather than something which depends on $N$. 
Your last three questions are related to applying the consequences of the first three.  
You cannot say $ \frac{(Np)!}{(Np-k)!} \to (Np)^{k}$  as $N \to \infty$ etc. even if that is what you want to say, so this approach does something more logically coherent
You could use Landau notation to say $ \frac{(Np)!}{(Np-k)!} \sim (Np)^{k}$ as $N \to \infty$ etc. and get to the result that way.  The answer you quote spells out the steps in a little more detail. 
A: The following fact is used three times: if $\frac{a}{b} \rightarrow 1$, then $a \rightarrow b$.
In particular, (2) is \begin{align*}
\frac{1}{(Np)^{k}}\frac{(Np)!}{(Np-k)!} &\xrightarrow{N\to\infty} 1 \\
\text{so} \qquad (A = {})\ \frac{(Np)!}{(Np-k)!} &\overset{N\to\infty}{\sim} (Np)^{k}
\end{align*}
and similarly for (3), giving an asymptotic value for $B$, and (4), giving an asymptotic calue for $C$.
In (5), $A$, $B$, and $C$ are replaced with their asymptotic values from (2), (3), and (4).
