Evaluating the limit $\lim_{n\to+\infty}(\sqrt[n]{n}-1)^n$

Question

Evaluate the limit $$\lim_{n\to+\infty}(\sqrt[n]{n}-1)^n$$

I know the limit is 0 by looking at the graph of the function, but how can I algebraically show that that is the limit?

Answer 1

$$a_n:=\left(\sqrt[n]n-1\right)^n\implies\log a_n=n\log\left(\sqrt[n]n-1\right)=-\infty\implies$$

$$\lim_{n\to\infty} a_n=\lim_{n\to\infty}e^{\log a_n}=0$$