= Solution
The <Hubble parameter> is $H=An t^{n-1}>0$, and
$$
\frac{\ddot a}{a}=An t^{n-2}\bigl[(n-1)+An t^n\bigr].
$$
For $n=1$ this is positive at every $t>0$, so expansion is exponential and inflationary throughout. For $0<n<1$, acceleration begins only after
$$
\boxed{t>t_*=\left(\frac{1-n}{An}\right)^{1/n}}.
$$
The interval $0<t<t_*$ is decelerating, while $t=t_*$ is the transition. Thus the model admits inflation, but not arbitrarily near its specified time origin when $n<1$; choosing parameters can place the transition early. Equivalently $\epsilon_H=(1-n)/(An t^n)$ becomes smaller than one at the same time.
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