= Solution
Let $D$ be the equal constant <density of states> per unit volume and energy in each band. In the <nondegenerate semiconductor> regime, the conduction-band occupation is
$$
f(E)\simeq e^{-(E-\mu)/(k_BT)}.
$$
For bands broad compared with $k_BT$, integrating above the conduction edge gives
$$
n=D\int_{E_c}^{\infty}e^{-(E-\mu)/(k_BT)}\,dE
=Dk_BT\,e^{-(E_c-\mu)/(k_BT)}.
$$
The valence-hole integral similarly gives $p=Dk_BT\,e^{-(\mu-E_v)/(k_BT)}$. Midgap $\mu$ therefore implies
$$
\boxed{n=p\simeq Dk_BT\,e^{-E_g/(2k_BT)}},\qquad E_g=E_c-E_v.
$$
Inserting the carrier densities into the transport expression yields
$$
\sigma\simeq\underbrace{eDk_BT(u_e+u_h)}_{\sigma_0(T)}
e^{-E_g/(2k_BT)}.
$$
\b[On treating the algebraic and mobility prefactor as approximately constant, the requested leading activation law is]
$$
\boxed{\sigma\propto e^{-E_g/(2k_BT)}}.
$$
The factor $1/2$ is essential: an intrinsic excitation creates an electron–hole pair, and charge neutrality places the <chemical potential> halfway across the symmetric gap. The conduction-electron population is governed by $E_c-\mu=E_g/2$, not by replacing this separation with the full gap.
There is a qualification to the proportionality as printed. A literal <constant-density-of-states intrinsic semiconductor> with temperature-independent mobilities retains the additional factor $T$. Indeed, using the full <Fermi-Dirac distribution> gives
$$
n=p=Dk_BT\ln\left(1+e^{-E_g/(2k_BT)}\right),
$$
which reduces to the expression above when $E_g\gg k_BT$. Thus the pure exponential is an activation approximation, not an exact consequence of constant DOS alone. A mobility sum proportional to $T^{-1}$ would cancel the leading algebraic factor.
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