sequences and series - Finding the limit $lim_{ntoinfty} frac{nleft(sqrt[n]{n}-1right)}{log n}$

I try to calculate the following limit:
$$\lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n}$$
I think it should equal 1, because:
$$\exp(x)=\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^{n}$$
(Already proven)

Solving for $x$ gives:
$$\log x = \lim_{n \to \infty} n \left(\sqrt[n]{x}-1\right) \implies \lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n}=\lim_{n\to\infty}\frac{\log n}{\log n}=1$$
But I like to calculate the limit with just standard things like L’Hôpital’s rule, because the previous way is maybe wrong and contains too much magic.

For example, I tried this:
$$\begin{align*} \lim_{n\to\infty}\frac{n\left(\sqrt[n]{n}-1\right)}{\log n} &= \lim_{n\to\infty}\frac{\frac{d}{dn}\!\!\left(n\left(\sqrt[n]{n}-1\right)\right)}{\frac{d}{dn}\log n} \\[6pt] &=\lim_{n\to\infty}\frac{n^{1/n-1}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n}-1}{1/n} \\[6pt] &=\lim_{n\to\infty}\left(n\left(n^{1/n-1}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n}-1\right)\right) \\[6pt] &=\lim_{n\to\infty}\left(n^{1/n}\left(-\left(\log\left(n\right)-1\right)\right)+n^{1/n+1}-n\right) \\[6pt] &=??? \end{align*}$$
But then it becomes really ugly and looks wrong. Have I done something wrong? Is there another way to find the limit?

Thank you for any ideas.

Answer

rewrite your term in the form $\frac{n^{1/n}-1}{\log(n)}{n}$ and use L'Hospital the result is $1$