= Solution
The <slow-roll approximation> neglects $\ddot\phi$ relative to $3H\dot\phi$ and kinetic energy relative to the potential. It is self-consistent on the attractor when the <potential slow-roll parameter> $\epsilon\ll1$ and the <second potential slow-roll parameter> $|\eta|\ll1$. Thus
$$
3H\dot\phi\simeq-V',\qquad H^2\simeq\frac{V}{3M_{\rm pl}^2},\qquad \epsilon=\frac{M_{\rm pl}^2}{2}\left(\frac{V'}V\right)^2,\qquad \eta=M_{\rm pl}^2\frac{V''}V.
$$
Here, as in the <inflaton> equations, $M_{\rm pl}$ denotes the <reduced Planck mass>, not the unreduced mass used in the thermal calculation. The <slow-roll curvature power spectrum> becomes
$$
\Delta_{\mathcal R}^2=\frac{H^4}{4\pi^2\dot\phi^2}\simeq\frac{9H^6}{4\pi^2V'^2}=\frac{V^3}{12\pi^2M_{\rm pl}^6V'^2},
$$
so
$$
\boxed{\Delta_{\mathcal R}^2(k)\simeq\left.\frac{V}{24\pi^2M_{\rm pl}^4\epsilon}\right|_{k=aH}.}
$$
To differentiate with respect to horizon-exit scale, write $\mathcal N=\ln a$, increasing with physical time. Along the slow-roll trajectory,
$$
\frac{d\phi}{d\mathcal N}=\frac{\dot\phi}{H}\simeq-M_{\rm pl}^2\frac{V'}V,\qquad\frac{d\ln k}{d\mathcal N}=1-\epsilon_H\simeq1-\epsilon.
$$
The difference between differentiating with respect to $\ln k$ and $\mathcal N$ contributes only at second slow-roll order to the tilt. Since $\ln\Delta_{\mathcal R}^2=3\ln V-2\ln|V'|+\mathrm{constant}$,
$$
\begin{aligned}
n_s-1&\simeq-M_{\rm pl}^2\frac{V'}V\left(3\frac{V'}V-2\frac{V''}{V'}\right)\\
&=-3M_{\rm pl}^2\left(\frac{V'}V\right)^2+2M_{\rm pl}^2\frac{V''}V.
\end{aligned}
$$
Therefore the <scalar spectral index in potential slow-roll parameters> is
$$
\boxed{n_s\simeq1-6\epsilon+2\eta.}
$$
All background quantities in these expressions are evaluated when the particular mode exits the <Hubble radius>.
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