= Solution
The <sequential-enzyme Hill limit> uses the intended <distinguished limit> for $n\geq2$ holds the other rates fixed while $k_1\to0$, $k_n\to\infty$, and $k_1k_n$ tends to a positive finite constant. Then for every $1\leq j<n$, $\phi_j=\prod_{i=1}^jk_i/(k_{-i}+k_i^+)=O(k_1)\to0$, whereas $\phi_n$ remains finite and nonzero. Hence intermediate complexes have negligible occupancy relative to the unbound and fully bound states.
In the <quasi-steady-state approximation> rate, all intermediate numerator and denominator terms disappear. Multiplying by $1/\phi_n=\prod_{i=1}^nK_i$ gives
$$
\boxed{\rho\simeq E_0\frac{k_n^+S^n}{\prod_{i=1}^nK_i+S^n}.}
$$
This is the <Hill equation> with maximal rate $E_0k_n^+$ and effective half-saturation <concentration> $(\prod K_i)^{1/n}$. Here its <power> $n$ follows from the sequential binding mechanism and the suppression of intermediate states.
The phrase “$k_1k_n$ finite” needs this scaling interpretation for a nontrivial limiting rate. Merely bounding that product, while permitting unrelated rates to vary, does not ensure that all intermediate weights vanish relative to the terminal one. The derivation explicitly supplies the usual fixed-other-rates, positive-product assumptions. For $n=1$, the rate in the box is already the exact one-state <quasi-steady-state approximation> rate, without this two-endpoint limiting argument.
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