= Solution
For the <strong Hill autorepression response>, in the requested approximation, $\dot Y=\beta K^nY^{-n}-\alpha Y$. Put $z=Y^{n+1}$. Then
$$
\dot z=(n+1)(\beta K^n-\alpha z),\qquad
z(t)=\frac{\beta K^n}{\alpha}[1-e^{-(n+1)\alpha t}],
$$
so the formal zero-initial-concentration solution is
$$
\boxed{Y(t)=\left(\frac{\beta K^n}{\alpha}\right)^{1/(n+1)}
[1-e^{-(n+1)\alpha t}]^{1/(n+1)},\qquad
Y_{\rm st}=\left(\frac{\beta K^n}{\alpha}\right)^{1/(n+1)}.}
$$
Setting $Y(T_{1/2})=Y_{\rm st}/2$ gives
$$
\boxed{T_{1/2}=\frac{-\log(1-2^{-(n+1)})}{(n+1)\alpha},\qquad
\frac{T_{1/2}}{T_{1/2}^{(0)}}=
\frac{-\log(1-2^{-(n+1)})}{(n+1)\log2}.}
$$
The formal ratio decreases with positive $n$; for $n=1$ it is about $0.208$, and for $n=2$ about $0.0642$. Its large-$n$ asymptote is $2^{-(n+1)}/[(n+1)\log2]$. This is a <Nonlinear ordinary differential equation>[nonlinear] half-response time, not simply the near-equilibrium <Time constant>[relaxation time] $1/[(n+1)\alpha]$.
\b[Approximation qualification.] The omitted initial layer is precisely the one the source permits us to ignore: the approximate derivative diverges at $Y=0$, whereas the full <Hill equation> has initial slope $\beta$. The approximation is not uniform as $n\to\infty$ at fixed $R=\beta/(\alpha K)$. Strong repression already at half of the final <concentration> needs $R^{1/(n+1)}/2\gg1$. The actual <Hill equation> obeys $\dot Y\le\beta$, so $T_{1/2}\ge Y_{\rm st}/(2\beta)$. Its steep-switch limit approaches the logic-response scale in <1/a>; the formal vanishing large-$n$ result above cannot be extrapolated past its regime of validity.
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