= Solution
If $n=\prod_{p^\alpha\parallel n}p^\alpha$, multiplicativity and the prime-power formula for the <divisor function> give
$$
\frac{\tau(n)}{n^\epsilon}
=\prod_{p^\alpha\parallel n}\frac{\alpha+1}{p^{\alpha\epsilon}}.
$$
Choose $P_0=2^{1/\epsilon}$. For $p\geq P_0$ and $\alpha\geq1$, the elementary bound $\alpha+1\leq2^\alpha$ gives
$$
\frac{\alpha+1}{p^{\alpha\epsilon}}\leq1.
$$
For each of the finitely many primes $p<P_0$, the sequence $(\alpha+1)p^{-\alpha\epsilon}$ tends to zero and therefore has a finite maximum $M_p$ over $\alpha\geq0$. Consequently
$$
\frac{\tau(n)}{n^\epsilon}
\leq C(\epsilon):=\prod_{p<P_0}M_p<\infty,
$$
uniformly in $n$, which proves $\tau(n)\leq C(\epsilon)n^\epsilon$.
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