Bound on Divisor Counting Function Let $d(n)$ be the number of divisors of n.
I'm trying to figure out if there is a constant C such that
\begin{equation}
d(n) \leq (ln(n))^C  
\end{equation}
My guess is there is no such constant. I have been trying to prove this by contradiction but haven't managed to do so. Any help would be appreciated!
 A: To look at whether or not $d(n)<(\log n)^C$ for some $C$, we want to investigate 
$$
\frac{\log d(n)}{\log \log n}.$$
One can show that 
$$
d(n) > \exp((1-\epsilon) \log 2 \cdot \log n/ \log \log n)
$$
for all $\epsilon >0$ and for infinitely many $n$.
Then 
$$
\log d(n) > (1-\epsilon) \log 2 \cdot \log n/ \log \log n
$$
for all $\epsilon >0$ and for infinitely many $n$.
Hence,
$$
\frac{\log d(n)}{\log \log n} > (1-\epsilon) \log 2 \cdot \log n/( \log \log n)^2$$
for all $\epsilon >0$ and for infinitely many $n$.
The right-hand side tends to infinity with $n$.  Hence, no such $C$ exists.
Reference: Gerald Tenenbaum, Introduction to Analytic and Probabilistic Number Theory, p. 82.
A few illustrating computations.  Let $p(m)$ be the product of the first $m$ primes.
Let $g(m)=\log(d(p(m)))/\log \log(p(m))$.  
Then $g(10)=2.223...$, $g(100)=11.132...$, $g(1000)=77.33..$.
Added: To get a lower bound on $d(n)$, let $n_k$ be the product of the first $k$ primes. Then $d(n_k)=2^k$ and
$$
\log n_k \le k \log p_k$$ where $p_k$ is the $k$-th prime.
By Chebyshev's estimate, we have
$$
\log n_k = \theta(p_k) \ge M p_k$$ for some positive constant $M$. 
Then we have
$$
\log d(n_k) \ge \frac{ \log 2 \log n_k}{\log p_k} 
\ge \frac{\log 2 \log n_k}{\log M + \log \log n_k}$$
from which we conclude that 
$$
\frac{\log n_k}{\log \log n_k} \rightarrow \infty$$ as $k \rightarrow \infty$.
