Let, $$\tau(n)=\sum_{d|n}1$$ Be the divisors counting function. Then is it true that, There exists infinitely many $n$ satisfying, $$\tau(n)>\left(\ln(n)\right)^{a}$$ Where $a\in[1,\infty)$?
My attempt:-
Let, $a=1$ Then all the integers of the form $n=2^{k}$ satisfy $$\tau(2^{k})>\ln(n)$$
But I was unable to do the same for the general case for $a$
Choose any positive integer $j \gt a$, and a set of distinct primes $p_i$ for $1\le i\le j$ (e.g., the smallest $j$ primes). Then, for any positive integer $k$, have
$$n = \prod_{i=1}^{j}p_i^k \;\;\to\;\; \tau(n) = (k+1)^{j} \tag{1}\label{eq1A}$$
Also, define
$$s = \sum_{i=1}^{j}\ln(p_i) \;\;\to\;\; \ln(n) = ks \tag{2}\label{eq2A}$$
We then have
$$\begin{equation}\begin{aligned} k & \gt s^{\frac{a}{j-a}} \\ k^{j-a} & \gt s^{a} \\ k^{j} & \gt k^{a}s^{a} \\ (k+1)^{j} & \gt (ks)^{a} \\ \tau(n) & \gt (\ln(n))^{a} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This shows that for all positive integers $k$ which satisfy the first line of \eqref{eq3A}, the inequality in the last line is also true. Thus, this proves there are infinitely many $n$ as defined in \eqref{eq1A} where the inequality at the end of \eqref{eq3A} is true for any specified $a \in [1,\infty)$.
