Estimate for the product of primes less than n In this paper Erdős shows a shorter proof for one of his old results stating that $$ s(n) = \prod_{p < n} p < 4^n$$ where the product is taken over all primes less than $n$. He also remarks that using the prime number theorem one can show $$ s(n)^{\frac1n} \stackrel{n\to\infty}{\longrightarrow} e.$$
Can someone here prove this result? It does not seem straightforward to me. 
One (crude) attempt I tried was to consider the product $$\prod_{i=2}^n \frac{i}{\log{i}} = n!\prod_{i=2}^n \frac{1}{\log{i}}$$ which I do not know how to estimate, not to mention that I would then have to argue that it is an asymptotic estimate for $s(n).$ 
Is there a simple way to show the result about $s(n)$ using the prime number theorem?

The summation formula at the top of Answer 1 isn't good necessarily percentage-wise.
For instance, if n = 14, the formula gives the estimate for the sum at about 50.6, but the sum of all of the primes from 2 to 13, inclusive, is actually 41.
