= Solution
For the <monomial inflation potential> on $\phi>0$, $V_{,\phi}/V=n/\phi$ and $V_{,\phi\phi}/V=n(n-1)/\phi^2$. Thus, to leading <slow-roll approximation>,
$$
\epsilon\simeq\frac{n^2M_{\rm Pl}^2}{2\phi^2},\qquad
\eta_V=\frac{n(n-1)M_{\rm Pl}^2}{\phi^2},\qquad
\eta\simeq4\epsilon_V-2\eta_V=\frac{2nM_{\rm Pl}^2}{\phi^2}.
$$
For $n=2,3,4$, $2n\geq n^2/2$. Therefore the loose order-one requirements $\epsilon\leq1$ and $|\eta|\leq1$ give
$$
\boxed{\phi^2/M_{\rm Pl}^2\geq2n.}
$$
This is the interpretation under which the printed threshold follows. It is not a sufficient criterion for genuine slow roll, which requires much smaller parameters and hence $\phi^2/M_{\rm Pl}^2\gg2n$ for these fixed powers. Nor is its equality the actual end of acceleration for $n=2,3$: there $\epsilon\simeq n/4<1$. For $n=4$ the leading estimate has $\epsilon\simeq1$. The potential-based order-one criterion $|\eta_V|<1$ also differs from the Hubble-flow criterion, especially at $n=4$. We use the endpoint the paper explicitly stipulates, rather than treating this boundary as a controlled slow-roll approximation.
Let $\phi_e^2=2nM_{\rm Pl}^2$ by that prescription. The <slow-roll e-fold count>, using $\dot\phi/H\simeq-nM_{\rm Pl}^2/\phi$, is
$$
N=\int_{t_i}^{t_e}H\,dt\simeq\frac1{M_{\rm Pl}^2}\int_{\phi_e}^{\phi_i}\frac{V}{V_{,\phi}}\,d\phi
=\frac{\phi_i^2-\phi_e^2}{2nM_{\rm Pl}^2}.
$$
For $N=60$, the <monomial inflation with a prescribed endpoint> result is
$$
\boxed{\phi_i^2/M_{\rm Pl}^2=122n,\qquad\epsilon_i\simeq n/244,\qquad\eta_i\simeq1/61.}
$$
Applying the leading <scalar spectral index in Hubble slow-roll parameters> and <tensor-to-scalar ratio> formulas gives
$$
\boxed{n_s\simeq1-\frac{n+2}{122},\qquad r\simeq\frac{4n}{61}.}
$$
These are predictions at the chosen 60-e-fold initial point, normally identified with the pivot scale's horizon exit. A physical endpoint determined by $\epsilon=1$ would instead have the leading estimate $\phi_e^2\simeq n^2M_{\rm Pl}^2/2$; its change to the e-fold count and predictions must not be silently mixed with the stipulated $122n$ calculation.
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