How to prove $\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\leq 2n$? I know that ($p$ prime)
(1) $$\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}\sim n$$
Is there a way to prove
(2) $$\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}< 2n$$
?
Thanks!
 A: Observe that, if for some sequence, we have $u_n \sim n$, that is $\left|\dfrac{u_n}{n}\right| \to 1$ then there exists $n_0$ such that  $\left|\dfrac{u_n}{n}\right| <2$ for all $n\geq n_0$. Thus we have $u_n<2n $ for all large $n$. 
You may apply it here with $u_n=\underset{p\leq n}{\prod}p^{\frac{1}{p-1}}$.
A: The following is a sketch. Using estimates, OlivierOloa's answer should give a proof that avoids the indefinite $n_0.$ The step at (*) is one suggestion (there may be easier ways) and I have omitted the calculation.   
In the Wiki entry for Prime Number Theorem there are estimates for the nth prime ascribed in footnotes to Dusart and Bach/Shallit. These bounds are sufficient to show the result you want. For $n\geq 6$ the bounds are:

$n(\log n+\log\log n-1)\leq p_n\leq n(\log n+\log\log n)$ 

If we take the log of both sides of $~~\prod_{p\leq n} p^{1/(p-1)}< 2n$ we get
$$\sum_{p\leq n} \frac{\log p}{(p-1)}< \log 2n .$$
The left side only increases at primes so it is enough to consider: 
$$\sum_{k=1}^{\pi(n)} \frac{\log p_k}{p_k-1} < \log 2p(\pi(n))=\log 2n.$$
Calling the lower estimate of $p~~ e_l$ and the upper estimate $e_u.$ Then the left side is maximally
$$\sum_{n=1}^{m}\frac{\log e_u}{e_l-1}. $$ 
And the r.h.s. is minimally $\log 2e_l$ evaluated at $n = m.$
The first terms in the sum should be computed directly as the estimate is not valid for these. 
*It is enough to show that the minimum of estimate for the r.h.s. and the maximum of estimate for the l.h.s. intersect at a certain point (between 5 and 10) and that the respective rates of change mean that the r.h.s will always exceed the l.h.s.
A plot (discrete points joined by lines for clarity) of the two estimates is shown below.

Then if we exhibit the inequality for the few primes not covered by the estimates we are done. 
