= Solution
In the nonrelativistic regime,
$$
E(p)=mc^2+\frac{p^2}{2m}+O\!\left(\frac{p^4}{m^3c^2}\right).
$$
The assumption $E-\mu\gg k_BT$ puts the <Fermi-Dirac distribution> in its <Maxwell-Boltzmann distribution>[Maxwell-Boltzmann limit], so
$$
\begin{aligned}
n&\simeq\frac{4\pi g_s}{h^3}
e^{(\mu-mc^2)/(k_BT)}
\int_0^\infty p^2e^{-p^2/(2mk_BT)}\,dp\\
&=\frac{4\pi g_s}{h^3}
e^{(\mu-mc^2)/(k_BT)}
\frac{\sqrt\pi}{4}(2mk_BT)^{3/2}.
\end{aligned}
$$
Therefore the <nonrelativistic Maxwell--Boltzmann number density> is
$$
\boxed{
n=g_s\left(\frac{2\pi mk_BT}{h^2}\right)^{3/2}
\exp\left(\frac{\mu-mc^2}{k_BT}\right).}
$$
For $p+e^-\leftrightarrow H+\gamma$, <chemical equilibrium> and the vanishing <photon chemical potential> imply
$$
\mu_p+\mu_e=\mu_H.
$$
Applying the number-density formula to each massive species and eliminating the chemical potentials gives
$$
\frac{n_en_p}{n_H}
=\frac{g_eg_p}{g_H}
\left(\frac{2\pi k_BT}{h^2}\right)^{3/2}
\left(\frac{m_em_p}{m_H}\right)^{3/2}
\exp\left(-\frac{(m_p+m_e-m_H)c^2}{k_BT}\right).
$$
Use charge neutrality, $n_p\simeq n_e$, the mass approximation $m_H\simeq m_p$, and the convention in the question that suppresses the order-one internal-degeneracy ratio $g_eg_p/g_H$. With the <Hydrogen binding energy>
$$
E_{\rm bind}=(m_p+m_e-m_H)c^2,
$$
we obtain the <Saha ionization equation>
$$
\boxed{
\frac{n_e^2}{n_H}\simeq
\left(\frac{2\pi m_ek_BT}{h^2}\right)^{3/2}
\exp\left(-\frac{E_{\rm bind}}{k_BT}\right).}
$$
The assumptions are <thermal equilibrium> and <chemical equilibrium> before <cosmological recombination>, a <nondegenerate gas>[nonrelativistic nondegenerate gas], zero <photon chemical potential>, <charge neutrality>, negligible proton-electron mass correction in the translational prefactor, and the stated <degeneracy factor>[degeneracy-factor approximation].
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