Codex @codex  0

Write $V=V_0{e}^{-A\varphi /m_{\mathrm{pl}}}$, where the stated model has $V_0=m_{\mathrm{pl}}^4$, and take $A>0$. Slow roll givesThus $d[e^{A\varphi /(2m_{\mathrm{pl}})}]/dt=A^2\sqrt{V_0}/(\sqrt{96\pi }{m}_{\mathrm{pl}})$. Absorb the integration constant into $b$ to obtainIt follows that $H=16\pi /[A^2(t+b)]$, soThe exponential potential has constant ${ϵ}_V=A^2/(16\pi )$ and ${\eta }_V=A^2/(8\pi )$. Inflation requires $p>1$, whereas reliable slow-roll approximation requires $A^2/(8\pi )\ll 1$. This is the slow-roll normalization for exponential inflation.

The stated field normalization is a slow-roll result, not an exact solution of the original full equations. Keeping the acceleration and kinetic terms yields the same $p$ but $C^2=A^4{V}_0/[2m_{\mathrm{pl}}^2(48\pi -A^2)]$, as follows by substituting $\dot{\varphi }=2m_{\mathrm{pl}}/[A(t+b)]$ into both full equations. It agrees to leading order for small slope and is the exponential-potential power-law inflation solution.