This problem is subordinate to finding out if $$\left(1+\frac{\log p_{n+1}}{p_n}\right)^{p_{n+1}/\log p_n},$$where $p_n$ is the $n$-th prime, never stabilizes above or below its limiting value, which is $e$. First of all, I tried using some bound for the $n$-th prime, but even the tightest I found (Dusart, 2010) didn't work.
So, here's the deal. Letting $x=p_{n+1}/\log p_n$ and $g(x)=\log p_{n+1}/p_n$ I considered the inequality $$\left(1+g(x)\right)^{x}<e\tag{$\star$}.$$ This and not the same in the other direction because even though initially $(\star)$ is most often false, and when it is true the difference in absolute value is smaller than when it is false, for $p_{n+1}/p_n$ small enough it is true (always for sufficiently large twin primes), and notoriously $\lim p_{n+1}/p_n =1$, hence I think it might change the trend, as $n\to\infty$; also because the classic $(1+1/x)^x$ approaches $e$ strictly from below. Indeed I noted that a successful strategy might be to find a family of functions $f$ maximizing the speed of convergence from below, since proving $g(x)<f(x)$ for large enough $x$ would result in solving my original problem. It is easy to see that $\frac{1}{x}$ is not what I'm looking for, since we trivially have $$\frac{\log p_{n+1}}{p_n}>\frac{\log p_n}{p_{n+1}}.$$ Soon I found what seems a good candidate, namely $$\left(1+\frac{\log(e+1/x)}{x\log\left(e-1/x^{5/3}\right)}\right)^x.$$ Below is its plot compared to that of $(1+1/x)^x$:
For $x>x_P\approx 3.0812$ the function is below $e$. And, in fact, the difference between the two functions does seem to be always greater than $\frac{1}x$.
Well, have I found a function belonging to the maximizing family? How can I accelerate even more the convergence from below? Thank you.