Solution
= Solution
Let $P=p_1p_2\cdots p_{k+1}$ be the product of the first $k+1$ <prime number>[primes], and take $n=P^a$ for positive integers $a$. The <prime factorization> gives
$$
\tau(P^a)=(a+1)^{k+1},
\qquad
(\log P^a)^k=a^k(\log P)^k.
$$
Their ratio is
$$
\frac{(a+1)^{k+1}}{a^k(\log P)^k}
=\frac{a(1+1/a)^{k+1}}{(\log P)^k}\longrightarrow\infty.
$$
\b[Thus every sufficiently large $a$ gives $\tau(P^a)\geq(\log P^a)^k$, producing infinitely many such $n$.]