calculus - Finding $limsup_{nrightarrowinfty} n^{frac{log(n)}{n}}$

I was attempting to show that the power series

\begin{equation*} \sum_{n=1}^\infty n^{\log(n)}z^n \end{equation*}

has a radius of convergence of $1$.

In order to do this I decided to use the $\alpha$ method. This meant evaluating the limit

\begin{equation*} \limsup_{n\rightarrow\infty} n^{\frac{\log(n)}{n}} \end{equation*}

I was able to prove that

\begin{equation*} \lim_{n\rightarrow\infty} \dfrac{\log(n)}{n} = 0 \end{equation*}

but I realized that was necessary but not sufficient to show that the limsup in question is $1$.

I then was able to prove that

\begin{equation*} \lim_{n\rightarrow\infty} n^{\frac{1}{n}} = 1 \end{equation*}

However I could not figure out any way of using that fact either.

I realized I could rewrite this limit as

\begin{equation*} \exp\left(\lim_{n\rightarrow\infty} \dfrac{\log(n)^2}{n}\right) \end{equation*}

However since I have not proven L'hôpital's rule, I have no way of evaluating this limit either.

This has left me pretty stuck. I'm not sure how I could tackle this problem from here. Where should I start for this limit? Is there a way to do it without L'hôpital's rule?

I'd really rather not know the whole proof if possible.