I'm not sure that I'm right.
First we have $\sum_{k=1}^n (n+1-k)/k = (n+1)H_n-n$, so $L = \lim_{n\to\infty} \left(\frac{(n+1)H_n-n}{\ln n!}\right)^{\frac{\ln n!}n}$ Take logarithm, we have $\ln L = \lim_{n\to\infty} A(n)B(n)$, where $A(n) = \frac{\ln n!}n = \frac{n\ln n+O(n)}n = \ln n+O(1)$ and $B(n) = \ln C(n)$ where \begin{align*} C(n) &= \frac{(n+1)H_n-n}{\ln n!} \\ &= \frac{(n+1)(\ln n+\gamma+O(1/n))-n}{n\ln n-n+O(\log n)} \\ &= \frac{n\ln n-(1-\gamma)n+O(\log n)}{n\ln n-n+O(\log n)} \\ &= \frac{1-\dfrac{1-\gamma}{\ln n}+O(1/n)}{1-\dfrac1{\ln n}+O(1/n)} \\ &= \left(1-\frac{1-\gamma}{\ln n}\right)\left(1-\frac1{\ln n}\right)^{-1}\left(1+O(1/n)\right)^2 \\ &= \left(1-\frac{1-\gamma}{\ln n}\right)\left(1+\frac1{\ln n}+O(1/\log n)^2\right)\left(1+O(1/n)\right) \\ &= 1+\frac\gamma{\ln n}+O(1/\log n)^2 \end{align*} So $B(n) = \ln C(n) = \ln\left(1+\frac\gamma{\ln n}\right)+O(1/\log n)^2 = \frac\gamma{\ln n}+O(1/\log n)^2$ and $A(n)B(n)=\gamma+O(1/\log n)$ Let $n\to\infty$, we have $\lim_{n\to\infty} A(n)B(n)=\gamma$, so $L = e^\gamma$.
The following equations come from Concrete Mathematics, proved by Euler-Maclaurin formula
- $H_n = \sum_{k=1}^n 1/k = \ln n+\gamma+O(1/n)$, where $\gamma$ is Euler-Mascheroni constant.
- $\ln n! = n\ln n-n+O(\log n)$. (It's really Stirling's approximation)