Compute $\lim _{n \rightarrow \infty}(\sqrt[n]{n}-1)^{\frac{1}{n}}$. 
I need to solve the following problem: $$ \lim _{n \mapsto
 \infty}(\sqrt[n]{n}-1)^{\frac{1}{n}}. $$

My attempt:
$$\lim _{n \mapsto
 \infty}\ln(\sqrt[n]{n}-1)^{\frac{1}{n}}=\lim _{n \mapsto
 \infty}\frac{\ln(\sqrt[n]{n}-1)}{n}.$$
Now I want to  prove that
$$\lim _{n \mapsto
 \infty}\frac{\ln(\sqrt[n]{n}-1)}{n}=0.$$
But I'm stuck on proving the limit without the Heine theorem (equivalence relation between limit of a sequence and limit of a function). Any help would be greatly appreciated.

Let $n$ be a positive integer.
Using Landau notations, $\sqrt[n]{n} -1= \exp(\frac{1}{n}\ln(n)) -1= \frac{\ln(n)}{n}+ o\left(\frac{\ln(n)}{n}\right)$ as $n\to+\infty$
Then $\frac{1}{n}\ln(\sqrt[n]{n}-1) = \frac{1}{n}\ln\left(\frac{\ln(n)}{n}+o\left(\frac{\ln(n)}{n}\right)\right) \to 0$ when $n\to+\infty$
And $(\sqrt[n]{n}-1)^\frac{1}{n}=\exp(\frac{1}{n}\ln(\sqrt[n]{n}-1)) \to \exp(0)=1$ when $n\to+\infty$
