Growth rate of product of smallest prime factors For $n\in \mathbb{N}$, let $p(n)$ denote the smallest prime dividing $n$. Then consider the function $f:\mathbb{N}\rightarrow \mathbb{N}$ defined by $f(n)= \prod_{k=1}^{n}p(k)$. 
Question: What is the growth rate of $f(n)$? Does it grow faster than any exponential function (i.e., any function of the form $a^{n}$ for some $a>1$)? 
Computing the asymptotic geometric mean of $f$ should essentially be equivalent to this question. That is, can one compute $\lim_{n\rightarrow \infty} f(n)^{1/n}$? I'm guessing the answer is $\infty$ but I haven't figured out a way of proving this. 
Thanks for reading.

$\displaystyle\pi(n) \sim \frac{n}{\ln n}$
The asymptotic expectation of $\displaystyle\frac{f(n+1)}{f(n)}$ is $n$ times the "asymptotic probability that n is prime", $\displaystyle\frac{n}{\ln(n)}-\frac{n-1}{\ln{n-1}}$, which is the expected number of primes $\leq n$ minus the expected number of primes $\leq n-1$. 
$\displaystyle\frac{f(n+1)}{f(n)}\sim n\left(\frac{n}{\ln(n)}-\frac{n-1}{\ln(n-1)}\right)=\frac{n^2[\ln(n-1)-\ln(n)]+n\ln(n)}{\ln(n)\ln(n-1)}$ which eventually grows like $\displaystyle\frac{n}{\ln(n-1)}$, which eventaully becomes greater than any $a$, hence the average growth rate is faster than exponential.
