RH would follow from $\displaystyle \frac{p_{n+1}}{p_{n+1}-1}<\frac{\log\log N_{n+1}}{\log\log N_n} $ for all $n>1$; what is my mistake? Let $N_n=\prod_{k=1}^np_k$ be the primorial of order $n$,$\gamma$ be the Euler-Mascheroni constant and $\varphi$ denote the Euler phi function. Nicolas showed that if the Riemann Hypothesis is true, then $$\forall \ n\in\mathbb{Z^+}, \ \frac{N_n}{\varphi(N_n)}=\prod_{k=1}^n\frac{p_k}{p_k-1}>e^\gamma \log\log N_n.\tag{NI}$$He also proved that the falsity of RH would imply the existence of infinitely many $n$ violating NI, as well as infinitely many $n$ satisfying it. Now, the combination of the following statement, equivalent to PNT:$$\lim_{n\to\infty}\frac{p_n}{\log N_n}=1$$ and Mertens' third theorem $$\lim_{n\to\infty}\frac{1}{\log p_n}\prod_{k=1}^n\frac{p_k}{p_k-1}=e^\gamma$$ yields $$\lim_{n\to\infty}\frac{1}{\log\log N_n}\prod_{k=1}^n\frac{p_k}{p_k-1}=e^\gamma,$$ thus NI would follow from the decreasingness, even only for large enough $n$, of the sequence $$u_n=\frac{1}{\log\log N_n}\prod_{k=1}^n\frac{p_k}{p_k-1}.$$ Therefore, since $\displaystyle \log \log 2 $ is negative, let us consider for $n>1$ $$\frac{u_{n+1}}{u_n}=\frac{\log\log N_n}{\log\log N_{n+1}}\frac{p_{n+1}}{p_{n+1}-1}<1 $$ $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log\log N_{n+1}}{\log \log N_n}=\frac{\log\log N_{n+1}}{\log(\log N_{n+1}-\log p_{n+1})}.\tag{$\star$}$$ 
We have $\log N_{n+1}<p_{n+1}$, wherefrom, for any real $r>\log p_{n+1}+1,$ $$\frac{1}{\log(r-\log\log N_{n+1})}<\frac{1}{\log(r-\log p_{n+1})},$$ so as long as $\log N_{n+1}>\log p_{n+1}+1$ for $n>1$, $(\star)$ is weaker than  $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log\log N_{n+1}}{\log(\log N_{n+1}-\log \log N_{n+1})}.\tag{1}$$
But $\displaystyle f(x)=\frac{\log x}{\log(x-\log x)}$ is decreasing on $(1,+\infty)$: $$\require\cancel f'(x)= \frac{\log(x-\log x)/x - \log x \cdot D(\log(x-\log x))}{\cancel{\log^2(x-\log x)}}<0 \\ \frac{\log(x-\log x)}{x}<\frac{(1-1/x)\log x}{x-\log x}=\frac{(x-1)\log x}{{x}(x-\log x)} \\ (x-\log x)\log(x-\log x)<(x-1)\log x \\ (x-\log x)^{x-\log x}<x^{x-1},$$ which follows from $$(x-\log x)^x < x^{x-1} \\ \left(1-\frac{\log x}{x}\right)^x<x^{-1} \\ \left(1-\frac{\log x}{x}\right)^{x/\log x}<x^{-1/\log x} =e^{-1},$$ which, setting $\displaystyle t=\frac{x}{\log x}$, reduces to the familiar $$\left(1-\frac{1}{t}\right)^t<e^{-1}.\tag{2}$$ With $x$ in place of $t$ we would say $(2)$ holds for $x>1$, but given that the range of $\displaystyle \frac{x}{\log x}$ does not include $[0,e)$, we really need $t>0$, or equivalently, again, $x>1$.
Hence $(1)$ is in turn implied by $$\frac{p_{n+1}}{p_{n+1}-1}<\frac{\log p_{n+1}}{\log(p_{n+1}-\log p_{n+1})} \\ p_{n+1}\log(p_{n+1}-\log p_{n+1})<(p_{n+1}-1)\log p_{n+1} \\ (p_{n+1}-\log p_{n+1})^{p_{n+1}}<p_{n+1}^{p_{n+1}-1},$$ and this we have already showed to ensue from $(2)$. Thence, $(\star)$ is true and so is NI.

However, not only such a result would prove RH, but in this paper it is shown that such a result would confute Cramér's conjecture. On account of this I expect to have made some stupid mistake, I just cannot see it. Thank you for any observations.
 A: The problem is the claim $\log\left(N_{n+1}\right)=\theta\left(p_{n+1}\right)<p_{n+1}
 $ where $\theta\left(x\right)$
  is the Chebyshev theta function. We knows the bounds $$n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2.050735}{\log\left(n\right)}\right)\leq\theta\left(p_{n}\right)\leq n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2}{\log\left(n\right)}\right)
 $$ and $$n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2.1}{\log\left(n\right)}\right)\leq p_{n}\leq n\left(\log\left(n\right)+\log\left(\log\left(n\right)\right)-1+\frac{\log\left(\log\left(n\right)\right)-2}{\log\left(n\right)}\right)
 $$ for every sufficiently large $n
 $ and these limitations don't guarantee your inequality.
A: Indeed, $\log N_{n+1}=\theta(p_{n+1})<p_{n+1}$ doesn't always hold. In particular, here is shown that $\theta(x)-x$ changes sign infinitely often. Since for $p_n<x<p_{n+1}$, $\theta(x)-x=\theta(p_n)-x<\theta(p_n)-p_n$, my inequality fails infinitely many times.
