How to prove $ \lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0 $. I am trying to prove that $ \lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0  $. My attempt is as follows.
Since for each $n\in \mathbb{N}$, $n>0$ by arithmetic geometric inequality $$ \sqrt[n]{n}=\sqrt[n]{n.1...1}=\le \dfrac{n+1+...+1}{n}=\dfrac{n+n-1}{n}=2-\dfrac{1}{n} $$.
Hence $$0<\sqrt[n]{n}-1\le 1-\dfrac{1}{n}<1.$$
So $ \lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0  $.
Is this correct ? If not how to show that $ \lim_{n\rightarrow \infty}\left( \sqrt[n]{n}-1 \right)^{n}=0  $ ? Please help me. Thanks.
 A: The inequality
$$ \sqrt[n]{n} \leq \frac{1+\ldots+1+n}{n}= 2-\frac{1}{n}$$
just gives:
$$ \limsup_{n\to +\infty}\left(\sqrt[n]{n}-1\right)^n \leq\frac{1}{e} $$
but if you consider the slightly improved inequality:
$$ \sqrt[n]{n}\leq\frac{1+\ldots+1+\sqrt{n}+\sqrt{n}}{n}=1-\frac{2}{n}+\frac{2}{\sqrt{n}}$$
you easily get:
$$ \lim_{n\to +\infty}\left(\sqrt[n]{n}-1\right)^n = 0 $$
as wanted.

For a sharp inequality, you need to write $n$ as a product of $n$ numbers, quite close one each other: that is not difficult to achieve. Since:
$$ n=1\cdot\prod_{k=1}^{n-1}\frac{k+1}{k} $$
by the AM-GM inequality it follows that:
$$ \sqrt[n]{n}\leq 1+\frac{H_{n-1}}{n} $$
that is essentially optimal since $e^x\geq 1+x$ implies:
$$ \sqrt[n]{n} \geq 1+\frac{\log n}{n}.$$
A: Not exactly rigorous, but you can make it rigorous without too much effort, I think: It suffices to show that
$$
\lim_{n \to \infty} \sqrt[n]{n} < 2
$$
But this is trivial, since $2^n \gg n$.  Hence the limit follows.  (In fact, $\lim_{n \to \infty} \sqrt[n]{n} = 1$.)
ETA: If the $2$ makes you nervous, you can substitute any value $q, 1 < q < 2$.  @Did chose $q = 3/2$ in the comments.
A: We can also approach this using brute force via application of L'Hospital's Rule.
$$\lim_{n\to \infty}(n^{1/n}-1)^n=\exp{\lim_{n\to \infty}n\log(n^{1/n}-1)}$$
Now, let's look at the interior limit.
$$\lim_{n\to \infty}n\log(n^{1/n}-1)=\lim_{n\to \infty}\frac{\log(n^{1/n}-1)}{n^{-1}}=-\lim_{n\to \infty}n^2\frac{1-n^{-2}\log n}{n^{1/n}-1}=-\infty$$
Inasmuch as $\lim_{x\to -\infty}e^x=0$, we obtain the expected result.
