= Solution
At horizon exit $k=aH$, $d\log k/dt=H(1-\epsilon_H)$, where $\epsilon_H=-\dot H/H^2$ is the <Hubble slow-roll parameter>. To leading slow-roll order,
$$
\frac{d}{d\log k}\simeq\frac{\dot\Phi}{H}\frac{d}{d\Phi}
\simeq-M_{\mathrm{Pl}}^2\frac{V'}V\frac{d}{d\Phi}.
$$
The <slow-roll curvature power spectrum> is proportional to $V^3/V'^2$. Therefore the <scalar spectral index in potential slow-roll parameters> is
$$
n_s-1\simeq-M_{\mathrm{Pl}}^2\frac{V'}V\left(3\frac{V'}V-2\frac{V''}{V'}\right)
=-3M_{\mathrm{Pl}}^2\left(\frac{V'}V\right)^2+2M_{\mathrm{Pl}}^2\frac{V''}V.
$$
Using the <potential slow-roll parameter> and the <second potential slow-roll parameter> gives
$$
\boxed{n_s\simeq1-6\epsilon_V+2\eta_V.}
$$
The neglected horizon-exit derivative correction contributes at higher slow-roll order.
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