= Solution
<Cosmic inflation> addresses the <horizon problem> by extending the causal past of the observed region, the <flatness problem> by damping the curvature contribution, and the <cosmological monopole problem> by diluting unwanted relics. A canonical <inflaton> can drive this expansion when its <potential energy> density exceeds twice its <kinetic energy> density.
Use the unreduced <Planck mass> $m_P=G^{-1/2}$, consistent with the coefficient in the <Friedmann equation>, and take $V_0>0$, $c\ne0$. Seek the <exponential-potential power-law inflation> solution $\phi=A\log(\beta t)$ and $a=a_*(t/t_*)^p$. The <kinetic energy> density is proportional to $t^{-2}$. The <inflaton potential> must have the same scaling, so
$$
\frac{cA}{m_P}=2,\qquad A=\frac{2m_P}{c},\qquad V(\phi(t))=\frac{V_0}{\beta^2t^2}.
$$
Substitution into the <inflaton equation of motion> gives
$$
-A+3pA-\frac{cV_0}{m_P\beta^2}=0,
\qquad\frac{V_0}{\beta^2}=\frac{2m_P^2}{c^2}(3p-1).
$$
The <Friedmann equation> now reduces to
$$
p^2=\frac{8\pi}{3m_P^2}\left(\frac{2m_P^2}{c^2}+\frac{2m_P^2}{c^2}(3p-1)\right)=\frac{16\pi p}{c^2}.
$$
Consequently the nontrivial positive-potential solution is
$$
\boxed{p=\frac{16\pi}{c^2},\qquad
\phi(t)=\frac{2m_P}{c}\log(\beta t),\qquad
\beta^2=\frac{V_0c^2}{2m_P^2(3p-1)},\qquad
a(t)=a_*\left(\frac{t}{t_*}\right)^p.}
$$
A positive $V_0$ requires $p>1/3$. <Accelerating expansion> occurs exactly for $p>1$, giving
$$
\boxed{c^2<16\pi.}
$$
This is an exact solution, not a use of the <slow-roll approximation>. The <first Hubble slow-roll parameter> and <equation-of-state parameter> are constant:
$$
\epsilon_H=-\frac{\dot H}{H^2}=\frac1p=\frac{c^2}{16\pi},\qquad
w=-1+\frac{2}{3p}=-1+\frac{c^2}{24\pi}.
$$
If the <reduced Planck mass> $M=m_P/\sqrt{8\pi}$ is used instead, the dimensionless exponential slope is $\lambda=c/\sqrt{8\pi}$ and the exponent becomes $p=2/\lambda^2$; these are the same solution.
For $p>1$, the <comoving Hubble radius> decreases as $(aH)^{-1}\propto t^{1-p}$. A region initially in causal contact can grow beyond that radius before the hot <radiation-dominated universe> begins. In the ideal power law extrapolated all the way to $t=0$, the <particle horizon> integral $\int_0^t dt'/a(t')$ diverges, so the solution has no finite past particle horizon. A finite, physically trustworthy inflationary interval instead requires a sufficiently large <number of e-folds> to enlarge an initially causal patch.
For a small curvature contribution, the <Friedmann equation> gives $|\Omega-1|=|k|/(a^2H^2)\propto t^{2-2p}\to0$, solving the <flatness problem>. Conserved nonrelativistic <magnetic monopoles> have <number density> proportional to $a^{-3}$, while the inflaton <energy density> is proportional to $t^{-2}$; their energy-density ratio therefore falls as $t^{2-3p}$. This gives the required dilution, provided later <reheating> does not recreate them. <Inflationary shear damping> similarly suppresses anisotropic expansion.
The main failure is \b[no automatic graceful exit or reheating]. The <first Hubble slow-roll parameter> never reaches one, and the exponential <inflaton potential> has no finite minimum about which the field can oscillate. Additional dynamics are needed for a <graceful exit from inflation> and transfer to a hot <radiation-dominated universe>. The ideal solution also retains a curvature singularity at $t=0$; solving the <horizon problem> does not itself explain the beginning of the universe. For $c=0$, the constant <inflaton potential> instead permits constant-field <de Sitter spacetime> expansion; the logarithmic field solution is inapplicable.
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