Evaluate $\lim_n \dfrac{\sqrt[n]{n!}}{n}$

Prove that $$\lim_{n\to \infty} \dfrac{\sqrt[n]{n!}}{n}=e^{-1}.$$

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I: Taking logarithm
Since $\ln \dfrac{\sqrt[n]{n!}}{n}=\ln \sqrt[n]{\dfrac{n!}{n^n}}=\dfrac{1}{n}\bigg(\ln \dfrac{1}{n}+\ln \dfrac{2}{n}+\cdots+\ln \dfrac{n}{n}\bigg)$, we have $$\begin{align}\lim_{n\to \infty} \dfrac{\sqrt[n]{n!}}{n} &=\exp\bigg(\lim\limits_{n\to \infty}(\dfrac{1}{n}\sum\limits_{k=1}^n\ln \dfrac{k}{n})\bigg)\\ &=\exp\bigg(\int_0^1 \ln x\;dx\bigg)\\ &=e^{-1}\;\Box\end{align}$$
II: Sandwich theorem
For any sequence $(a_n)$ of positive real numbers, we have $$\liminf_n \dfrac{a_{n+1}}{a_n}\leq \liminf_n a_n^{1/n}\leq \limsup_n a_n^{1/n}\leq \limsup_n \dfrac{a_{n+1}}{a_n}$$ Pick $a_n=\dfrac{n!}{n^n}$. Then $\dfrac{a_{n+1}}{a_n}=(1+\dfrac{1}{n})^{-n}$ and $a_n^{1/n}=\dfrac{\sqrt[n]{n!}}{n}$. It follows that $$\liminf_n (1+\dfrac{1}{n})^{-n}\leq \liminf_n \dfrac{\sqrt[n]{n!}}{n}\leq \limsup_n \dfrac{\sqrt[n]{n!}}{n}\leq \limsup_n (1+\dfrac{1}{n})^{-n}.$$ We know $$\liminf_n (1+\dfrac{1}{n})^{-n}=\limsup_n (1+\dfrac{1}{n})^{-n}=e^{-1}.$$ By sandwich theorem, we have $$\liminf_n \dfrac{\sqrt[n]{n!}}{n}=\limsup_n \dfrac{\sqrt[n]{n!}}{n}=\lim_n \dfrac{\sqrt[n]{n!}}{n}=e^{-1}\;\Box$$
III: Stirling approximation $$n!\sim \sqrt{2\pi n}(\dfrac{n}{e})^n\\ \dfrac{\sqrt[n]{n!}}{n}\sim \dfrac{\sqrt[2n]{2\pi}e^{-1}n^{1+1/(2n)}}{n}\to e^{-1}\;\Box$$
IV: Lemma: If $\lim_{n\to\infty}a_n=a$ and $a_n>0$ for all $n$, then we have $$ \lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a.$$ Pick $a_n=\bigg(1+\dfrac{1}{n}\bigg)^n$. Since $a_n>0$ for all $n$ and $\lim_{n\to \infty}a_n=e$, we can apply the lemma: $$\begin{align} e&=\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}\\ &=\lim_{n\to\infty}\sqrt[n]{\left(\frac{2}{1}\right)^1\left(\frac{3}{2}\right)^2\cdots\left(\frac{n+1}{n}\right)^n}\\ &=\lim_{n\to\infty}\sqrt[n]{\frac{(n+1)^n}{n!}}\\&= \lim_{n\to\infty}\frac{n+1}{\sqrt[n]{n!}}\\&=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}+\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}} \tag{*}\\&=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}},\end{align}$$ where in $(*)$ we have also used the lemma for $a_n=\dfrac{1}{n}\to 0$. It follows that $$\lim_{n\to \infty} \dfrac{\sqrt[n]{n!}}{n}=e^{-1}\;\Box$$
Reference
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