Let $p_1 < p_2 <\dots < p_n$ be the $n$ first primes listed in crescent order. Using the Chebyshev Inequality (for $x$ sufficiently large)
$$0.92\leq \frac{\pi(x)\log x}{x}\leq 1.11,$$
How to prove that, given a $\varepsilon > 0, $ there is a $M >0$, such that
$$0.92(1-\varepsilon)\leq \frac{n\log n}{p_n}\leq 1.11(1+\varepsilon)?$$
For each $n\geq 1$, let be $M_n = p_1p_2\dots p_n$. Considering the Stirling inequality $$\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\leq n!\leq\sqrt{2\pi n}\left(\frac{n}{e}\right)^n e^{1/2},$$ how to find a bound for $\pi(M_n)$?
Since $\pi(p_n)=n$, the Chebyshev bound gives that for any $n$ big enough we have: $$ 0.92 \leq \frac{n\log(p_n)}{p_n} \leq 1.11 \tag{1}$$ hence: $$\frac{n\log(p_n)}{1.11}\leq p_n\leq\frac{n\log(p_n)}{0.92}\tag{2}$$ and $$\log n+\log \log p_n -\log 1.11 \leq \log(p_n)\leq \log n +\log \log p_n - \log 0.92\tag{3}$$ so $$\log(p_n) = \log n\cdot \left(1+O\left(\frac{\log \log n}{\log n}\right)\right)\tag{4}$$ and it is sufficient to plug $(4)$ into $(1)$ to prove the first claim.
However, using the Chebyshev bound and the Stirling's inequality does not provide good bounds for $\pi(M_n)$ - we get something like: $$ C_1 (0.92-\varepsilon)^n n^n e^{-n}\sqrt{2\pi n}\leq \pi(M_n) \leq C_2 (1.11+\varepsilon)^n n^n e^{-n}\sqrt{2\pi n}$$ where $C_1$ and $C_2$ are two constants depending on when the Chebyshev bound becomes effective.