Solution
= Solution
We first prove by induction that
$$
a_n\geq2^n\qquad(n\geq1).
$$
It holds for $n=1$, and if it holds at $n$, then
$$
a_{n+1}=2^{a_n}\geq2^{2^n}\geq2^{n+1}.
$$
Consequently
$$
0\leq\frac{2n^k}{a_n}\leq2^{1-n}n^k.
$$
Part (c) and the <squeeze theorem> now give
$$
\boxed{\lim_{n\to\infty}\frac{2n^k}{a_n}=0}.
$$