what is the intuition behind the SRSWOR formula? I earlier asked about Slovin's Formula (https://math.stackexchange.com/questions/1410492/what-is-the-intuition-behind-slovins-formula), and learned shortly thereafter that it was derived from this formula. 
$n=\dfrac{n_0}{1+\dfrac{n_0}{N}}$,
Where $n_0=\dfrac{z^2p(1-p)}{e^2}$,
$N$ is the population size, $z$ is the standard normal variate based on the confidence coefficient, $p$ is the estimate for $P$, and $e$ is a specified margin of error.
So, breaking it down, what factors take part in determining the sample size?
(read the comments for further context)
 A: The intuition is that you don't have infinity in the finite population world. You cannot have a sample more of size more than $N$. As $n_0 \to \infty$, for whatever reason (you keep making the specified margin of error smaller), $n\to N$ because there is no room to grow beyond $N$. The numerator and denominator may or may not represent anything in particular: you can multiply them both by $N$ or divide by $n_0$, which would change what is in the numerator and in the denominator... But the way the formula is written as is, $n_0$ is the required sample size from an infinite population, and the denominator is some sort of undoing the finite population correction (FPC), of SRSWOR sampling variance: $1-n/N$. Thinking again about large samples, as $n\to N$, the variance goes to zero, which denotes that you don't have any sampling error (you may still have measurement error, of course, but the finite population sampling paradigm assumes fixed and perfectly measured values, at least as the first natural science approximation to whatever complexities might be encountered in practice). The formula you gave undoes this FPC.
