Prove that $2^x < \prod_{p\le x} p < (13/4)^x$ for sufficiently large x. Here $p$ is prime.
So if we take logs we need to show for sufficiently large x, $x\log 2 < \sum_{p\le x}\log p < x\log(13/4)$. Also according to http://en.wikipedia.org/wiki/Primorial
Asymptotically, primorials ''pn#'' grow according to:
$p_n\# = e^{(1 + o(1)) n \log n}$
where $o(\cdot)$ is the little o notation.