From Nielsen & Chuang, page 55:
The basic intuition for this decrease in resources required can be understood quite easily. Suppose the source emitting states $|0\rangle$ with probability $p$ and $(|0\rangle + |1\rangle)/\sqrt2$ with probability $1 — p$ is used a large number $n$ times. Then by the law of large numbers, with high probability the source emits about $np$ copies of $|0\rangle$ and $n(1 — p)$ copies of $(|0\rangle + |1\rangle)/\sqrt2$. That is, it has the form $$|0\rangle^{\otimes np} \left( \frac{|0\rangle + |1\rangle}{\sqrt2} \right)^{\otimes n(1-p)},\tag{1.60}\label{1.60}$$ up to re-ordering of the systems involved. Suppose we expand the product of $|0\rangle + |1\rangle$ terms on the right hand side. Since $n(1 — p)$ is large, we can again use the law of large numbers to deduce that the terms in the product will be roughly one-half $|0\rangle$s and one-half $|1\rangle$s. That is, the $|0\rangle + |1\rangle$ product can be well approximated by a superposition of states of the form $$|0\rangle^{\otimes n(1-p)/2} |1\rangle^{\otimes n(1-p)/2}.\tag{1.61}\label{1.61}$$ Thus the state emitted by the source can be approximated as a superposition of terms of the form $$|0\rangle^{\otimes n(1+p)/2} |1\rangle^{\otimes n(1-p)/2}.\tag{1.62}\label{1.62}$$ How many states of this form are there? Roughly $n$ choose $n(1 + p)/2$, which by Stirling's approximation is equal to $N \equiv 2^{nH[(1+p)/2,(1-p)/2]}$. A simple compression method then is to label all states of the form (\ref{1.62}) $|c_1\rangle$ through $|c_N\rangle$. It is possible to perform a unitary transform on the $n$ qubits emitted from the source that takes $|c_j\rangle$ to $|j\rangle|0\rangle^{\otimes n-nH[(1+p)/2,(1-p)/2]}$, since $j$ is an $nH[(1+p)/2,(1-p)/2]$ bit number. The compression operation is to perform this unitary transformation, and then drop the final $n-nH[(1+p)/2,(1-p)/2]$ qubits, leaving a compressed state of $nH[(1+p)/2,(1-p)/2]$ qubits. To decompress we append the state $|0\rangle^{\otimes n-nH[(1+p)/2,(1-p)/2]}$ to the compressed state, and perform the inverse unitary transformation.
I don’t understand the emphasized part. How to obtain the result $N$ by Stirling’s approximation? I’m not familiar with Stirling’s approximation. Can anyone explain in detail?
The quantity that's being described, $n$ choose $n(1+p)/2$ (also known as a binomial coefficient), is given by $$ \begin{pmatrix} n \\ k \end{pmatrix} = \frac{n!}{k!(n-k)!} $$ where you take $k = \frac12 n(1+p)$. Stirling's approximation, which reads $$ n! \sim \sqrt{2\pi n} \frac{n^n}{e^n}, $$ can then be applied to the three factorials in the binomial coefficient. The rest is simple algebra and it is really up to you to implement.