How to show how primorials grow asymptotically? The primorial $p_n\# $ is defined as the product of the first $n$ primes:
$$p_n\# = \prod_{k = 1}^n p_k.$$
Asymptotically, primorials grow like
$$p_n\# = e^{(1 + o(1))n\ln n)}.$$
How does one derive this asymptotic formula?
 A: Let $\pi(x)$ denote the prime counting function.  Let $x\geq 1$.  We will compute $\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)$ where the sum is taken over all primes $p\leq x$.  Note that 
$\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)=\displaystyle\sum_{p\leq x\;prime}\int^x_p 1/t dt=\int_1^x \frac{\pi(t)}{t}dt$.  
Since there exists a constant $c$ such that $\pi(t)\leq \frac{ct}{\ln t}$, it follows that 
$\displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)=O(\int_1^x \frac{1}{\ln t} dt)=o(x)$.
The prime number theorem gives
$\displaystyle\sum_{p\leq x\;prime}\ln p=\pi(x)\ln x - \displaystyle\sum_{p\leq x\;prime} (\ln x - \ln p)= \ln x (\frac{x}{ln x}+o(\frac{x}{ln x}))-o(x)=x+o(x)$.
Then $\displaystyle\sum_{k=1}^n \ln p_k = \displaystyle\sum_{p\leq p_n\;prime}\ln p=p_n+o(p_n)$.  It is known that $p_n=n\ln n +o(n\ln n)$; see for example the Wikipedia article on the prime number theorem.  Hence $\displaystyle\sum_{k=1}^n \ln p_k =n\ln n + o(n\ln n)$.  Taking exponentials of both sides gives the result you asked for.
