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.

We have for n>2
$\frac{\ln(\sqrt[n]{n}-1)}{n} = \frac{\ln\left[\sqrt[n]{n}\left(1-\frac{1}{\sqrt[n]{n}}\right)\right]}{n}$
$= \frac{\ln(\sqrt[n]{n})}{n}+\frac{\ln\left(1-\frac{1}{\sqrt[n]{n}}\right)}{n}$
$= \frac{\ln(n)}{n^2}+\frac{\ln\left(1-\frac{1}{\sqrt[n]{n}}\right)}{n}$
But for x>1, $\ln(x)\leq x-1$ so
$0\leq \frac{\ln(n)}{n^2} \leq \frac{n-1}{n^2}$
and the sequence $\left(\ln\left(1-\frac{1}{\sqrt[n]{n}}\right)\right)_{n\geq 2}$ is bounded...
