How do I compare the growth rate of functions $f(n)=(2^{n+1}-1)^{1/\log n}$ and $g(n)=n^{\log n}$
My Attempt
\begin{align} \lim_{n\to\infty}\frac{\log f(n)}{\log g(n)}&=\lim_{n\to\infty}\frac{\log\Big(2^{n+1}-1\Big)^{1/\log n}}{\log 2^{(\log n)^2}}\\ &\approx\lim_{n\to\infty}\frac{\frac{1}{\log n}\log\big(2^{n+1}\big)}{{(\log n)^2}}\quad\Bigg[2^{n+1}-1=2^{n+1}\text{ for large }n\\ &=\lim_{n\to\infty}\frac{n+1}{(\log n)^3}\\ &=\lim_{n\to\infty}\frac{1}{3(\log n)^2.\frac{1}{n\ln 2}}\bigg[\text{Applying L'Hospital rule}\\ &=\lim_{n\to\infty}\frac{n\ln 2}{3(\log n)^2}\\ &=\lim_{n\to\infty}\frac{\ln 2}{6(\log n).\frac{1}{n\ln 2}}\\ &=\lim_{n\to\infty}\frac{(\ln 2)^2.n}{6\log n}=\infty\\ \end{align}
But when I tried to plot the graphs in Mathematica I am getting $g(n)$ to be larger than $f(n)$:
Note: $\log n=\log_2 n$ and $\ln 2=\log_e 2$
Mathematica code: $\text{Plot}\left[\left\{f(n)=\left(2^{n+1}-1\right)^{\frac{1}{\log _2(n)}},g(n)=n^{\log _2(n)}\right\},\{n,0,10\},\text{PlotLabels}\to \text{Expressions},\text{ImageSize}\to \text{Large}\right]$