$\lim_{n\to\infty}\left(\frac{\log(p_{n+1})}{\log(p_n)}\right)^n = C?$ There is a conjecture (which is weaker) related conjecture to Firoozbakht's conjecture (see OEIS A182514 Commments) which states (and define $L_n$): $$L_n := \left(\frac{\log(p_{n+1})}{\log(p_n)}\right)^n < e,$$ where $p_n$ is the $n$-th prime.
This implies there might be constant, $C$, because as the RHS is constant as $n\to\infty$, and every thing on the LHS increasing without bound. So does $C$ exist and what value does it approach?
While I know that the $\lim_{n\to\infty}\left(\frac{\log(p_{n+1})}{\log(p_n)}\right) = 1$, and clearly $\left(\frac{\log(p_{n+1})}{\log(p_n)}\right) > 1$, it is the power, $n$, that is a curve-ball for me. 

Does the limit exist, and if it does, is the value $1,e$, or something
  in between for: $$\lim_{n\to\infty} L_n = C?$$

 A: Based on what we believe to be true about primes, the limit in question doesn't exist; its lim inf is $1$ but its lim sup is greater than $1$, though probably finite. For simplicity we shall examine the logarithm of your expression,
$$
\log L_n = n \log \frac{\log p_{n+1}}{\log p_n}.
$$
Define a normalized gap between primes: $g_n = (p_{n+1}-p_n)/(\log p_n)^2$, so that $p_{n+1} = p_n + g_n \log^2 p_n = p_n (1+(g_n \log^2 p_n)/p_n)$. Then using $\log(1+x)=x+O(x^2)$ when $x=o(1)$, we have
\begin{align*}
n \log \frac{\log p_{n+1}}{\log p_n} &= n \log \frac{\log p_n +\log(1+(g_n \log^2 p_n)/p_n)}{\log p_n} \\
&= n \log \bigg( 1 + \frac{(g_n \log^2 p_n)/p_n + O((g_n \log^2 p_n)^2/p_n^2)}{\log p_n} \bigg) \\
&= n \bigg( \frac{g_n \log p_n}{p_n} + O\bigg( \frac{g_n^2 \log^3 p_n}{p_n^2} \bigg) \bigg).
\end{align*}
By the prime number theorem, $n = p_n/\log p_n+O(p_n/\log^2 p_n)$, so
\begin{align*}
n \log \frac{\log p_{n+1}}{\log p_n} &= \bigg( \frac{p_n}{\log p_n} + O\bigg( \frac{p_n}{\log^2 p_n} \bigg) \bigg) \bigg( \frac{g_n \log p_n}{p_n} + O\bigg( \frac{g_n^2 \log^3 p_n}{p_n^2} \bigg) \bigg) \\
&= g_n + O\bigg( \frac{g_n}{\log p_n} \bigg).
\end{align*}
If gaps between consecutive primes of size $x$ were $o(\log^2x)$ (which we currently cannot disprove), then $g_n$ would be $o(1)$ and hence $\log L_n$ would tend to $0$. Certainly the average gap is $\log x$, so $g_n\ll1/\log p_n$ must occur and the lim inf of $\log L_n$ equals $0$.
If gaps between consecutive primes of size $x$ were occasionally as large as $\Theta(\log^2x\log\log x)$ (which we also currently cannot disprove), then $g_n$ and hence $\log L_n$ would be unbounded above.
However, we believe that the maximal order of magnitude of prime gaps near $x$ is $\log^2x$. Therefore we believe that the lim inf of $g_n$ is positive and finite, which would say the same about $\log L_n$.
The actual lim sup of $g_n$ is a bit mysterious, but some now believe that its lim sup should equal $2e^\gamma$, where $\gamma$ is Euler's constant. That would imply that $\limsup L_n = e^{2e^\gamma}$ and hence disprove  Firoozbakht's conjecture.
A: Some extra notes.
$L_n := \left(\frac{\log(p_{n+1})}{\log(p_n)}\right)^n < e.$
$\frac{\log(p_{n+1})}{\log(p_n)} < e^{1/n},$ both sides approach 1.
$\frac{\log(p_{n+1})}{e^{1/n}\log(p_n)} = \frac{e^{-1/n}\log(p_{n+1})}{\log(p_n)} < 1,$ left side approaches 1.
