= Solution
In the nondegenerate <conduction band>, the <Fermi-Dirac distribution> reduces to its <Boltzmann distribution> tail. Thus
$$
\begin{aligned}
n&=\int_{\epsilon_c}^\infty g_e(\epsilon)e^{-(\epsilon-\mu)/(k_BT)}d\epsilon\\
&=N_c e^{-(\epsilon_c-\mu)/(k_BT)},\qquad
\boxed{N_c=2\left(\frac{m_e^*k_BT}{2\pi\hbar^2}\right)^{3/2}}.
\end{aligned}
$$
Here the substitution $x=(\epsilon-\epsilon_c)/(k_BT)$ leaves $\int_0^\infty x^{1/2}e^{-x}dx=\sqrt\pi/2$. For <electron holes> use the empty-state probability $1-f(\epsilon)\simeq e^{(\epsilon-\mu)/(k_BT)}$, not the conduction-band Boltzmann factor. This gives
$$
p=N_v e^{-(\mu-\epsilon_v)/(k_BT)},\qquad
\boxed{N_v=2\left(\frac{m_h^*k_BT}{2\pi\hbar^2}\right)^{3/2}}.
$$
The <chemical potential> cancels in the product. With $E_g=\epsilon_c-\epsilon_v$,
$$
\boxed{np=N_cN_v e^{-E_g/(k_BT)}
=\frac12(m_e^*m_h^*)^{3/2}\left(\frac{k_BT}{\pi\hbar^2}\right)^3e^{-E_g/(k_BT)}=n_i^2.}
$$
This <semiconductor mass-action law> holds in thermal equilibrium for nondegenerate carriers, including a doped <semiconductor>. Only in an <intrinsic semiconductor> must $n=p=n_i$. The factors $N_c,N_v$ are the <effective density of states>; the factor $1/2$ in the final expression follows from the two spin factors already included in (a).
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