How is $\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)}{\ln(n)}\left(1 + \frac{1}{n}\right) \right)^n = e$? Tried this question here How to calculate $\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)^{n+1}}{\ln n^n} \right)^n$? and was curious about the result. The answer according to Wolfram Alpha is $e$, so I wanted to try it.
$\lim\limits_{n \to \infty} \left( \frac{\ln((n+1)^{n+1})}{\ln (n^n)} \right)^n$
$\lim\limits_{n \to \infty} \left( \frac{(n+1)\ln(n+1)}{n\ln (n)} \right)^n$
$\lim\limits_{n \to \infty} \left( \frac{\ln(n+1)}{\ln(n)}\left(1 + \frac{1}{n}\right) \right)^n$
This is similar to the typical definition $\lim\limits_{n \to \infty} \left(1 + \frac{1}{n}\right)^n = e$ but it has the extra log factors.
How come these two happen to be equivalent? Is it valid to apply L'Hospital's Rule to the logs even though they're inside the $()^n$? Or can it be applied to just part of the function and not the other half? What's the correct way to handle this extra log multiplier?
For instance:
$\lim\limits_{n \to \infty}\frac{\ln(n+1)}{\ln(n)} = \lim\limits_{n \to \infty}\frac{\frac{d}{dn}\ln(n+1)}{\frac{d}{dn}\ln(n)} = \lim\limits_{n \to \infty}\frac{n}{1+n} = \lim\limits_{n \to \infty}\frac{1}{1/n+1}  = 1$
but I don't think we can necessarily analyze this "separately" from the main result; I think they must be taken together somehow.  I also considered squeeze theorem but couldn't think of another function approaching $e$ from the other side.
 A: With a Taylor expansion-based argument:
When $n\to\infty$, we get
$$
\frac{\ln(n+1)}{\ln n}= \frac{\ln n+\ln\left(1+\frac{1}{n}\right)}{\ln n}
= 1+ \frac{\ln\left(1+\frac{1}{n}\right)}{\ln n}
= 1 + \frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right) \tag{1}
$$
(using that $\ln(1+x)=x+o(x)$ when $x\to0$) so that
$$\begin{align}
\frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right) &= 
\left(1 + \frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\right)\left(1+\frac{1}{n}\right)
= 1+\frac{1}{n}+\frac{1}{n\ln n} + o\left(\frac{1}{n\ln n}\right)\\
&= 1+\frac{1}{n}+o\left(\frac{1}{n}\right) \tag{2}
\end{align}$$
and from (2) and the same Taylor expansion of $\ln(1+x)$ at $0$ we get
$$\begin{align}
\left(\frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right)\right)^{n}
&= e^{n\ln \left(\frac{\ln(n+1)}{\ln n}\left(1+\frac{1}{n}\right)\right)}
= e^{n\ln \left(1+\frac{1}{n}+o\left(\frac{1}{n}\right)\right)}
= e^{n\left(\frac{1}{n}+o\left(\frac{1}{n}\right)\right)}
= e^{1+o\left(1\right)} \\&\xrightarrow[n\to\infty]{} e^1 = e
\end{align}$$
as claimed.
A: Use my comment in the question mentioned to use that if $a_n\to a$ then $$\left(1+\frac{a_n}{n}\right)^n \to e^a$$ in this case $$a_n=n\frac{\ln(n+1)-\ln n}{\ln n}=\frac{1}{\ln  n}\ln \left(1+\frac{1}{n}\right)^n\to 0$$ and thus 
$$\left(\frac{\ln (n+1)}{\ln n}\right)^n\to 1$$
