Enquiry on primorial numbers and primes Does the inequality 
$$1 + \dfrac{\log p_{k+1}}{\log N_k} > (\log N_k)^{\dfrac{1}{p_{k+1} -1}}$$
hold for all integers $k\geq 2$, where $N_k$ denotes the $k$-th primorial number $N_k=p_1p_2\cdots p_k$ and $p_j$ the $j$-th prime number?
What I think: 
By the prime number theorem, we know that there exists some constant $$1/2 \leq \theta <1$$ such that $$\log N_k = p_k + O({p_{k}^{\theta}\log p_k})$$
and upon invoking this together with Erdos' result that $p_{k+1}$ tends to $p_k$ for large $k$, one notices that the RHS of the inequality converges faster with respect to the LHS, so that if it holds for some $k=k_{0}$, then it holds for all $k\geq k_{0}$. Taking $k_{0} = 2$, we obtain the desired result ?
 A: The answer is probably no. Your inequality implies
$$ \frac{\log p_{k+1}}{\log N_k} > \log\left(1+\frac{\log p_{k+1}}{\log N_k}\right)>\frac{\log\log N_k}{p_{k+1}-1}>\frac{\log\log N_k}{p_{k+1}}.$$
In particular,
$$ p_{k+1}\log p_{k+1}>\theta(p_k)\log\theta(p_k),$$
whence
$$ p_{k+1}>\theta(p_k). $$
By Littlewood's theorem, the right hand side exceeds $p_k+p_k^{1/2}$ infinitely often, while it is widely believed that the left hand side is smaller than $p_k+p_k^{1/3}$ for every large $k$. So, if we believe the last upper bound, there are infinitely many counterexamples. By quoting Littlewood's theorem a bit more precisely, we get a contradiction already by Legendre's conjecture that there is a prime number between any two large consecutive squares.
A: Not an answer yet.
As in my comment, we may rewrite the inequality as $\log N_{k+1}>\left(\log N_k\right)^{\frac{p_{k+1}}{p_{k+1}-1}}$. Introduce the function $f_a(x)=\log(ax)-\left(\log a\right)^{\frac{x}{x-1}}$ so that the problem amounts to $f_{N_k}(p_{k+1})>0$. Notice 
$$\frac{d}{dx}f_a(x)=\frac1x+\frac{(\log a)\,a^{\frac{x}{x-1}}}{(x-1)^2}>0$$
(of course $x, a>1$) shows $f_a(x)$ is an increasing function of $x$. That means, once $x$ is large enough we conquer positivity as the OP noted.
I claim a stronger inequality $f_{N_k}(p_k)>0$ which improves $f_{N_k}(p_{k+1})>0$.
