Find $\lim_{ n \to \infty} (\frac{n!}{n^n})^{\frac{1}{n}}$ Find the limit : 
$\lim_{ n \to \infty} (\frac{n!}{n^n})^{\frac{1}{n}}$
My working : 
Let $$t = \lim_{ n \to \infty} (\frac{n!}{n^n})^{\frac{1}{n}}$$
Now taking log on both sides : 
$$\log t = \frac{1}{n}\log \frac{n!}{n^n}$$ ( we will consider the limit later on ) 
$$\log t = \frac{1}{n}( \log n! - \log n^n)$$ 
$$\log t = \frac{1}{n}\log n! -\frac{n}{n} \log n $$
$$\log t = \lim_{ n \to \infty} \frac{1}{n}\log n! - \lim_{ n \to \infty} \log n $$
$$\log t = 0 - \lim_{ n \to \infty} \log n $$
I know its duplicate problem at this site but I want to do it in more easier way by taking logs rather using Cauchy-d'Alembert criterion, Stolz–Cesàro theorem, or The Lalescu sequence
Can you please help me from here... thanks... 
 A: STolz theorem
$$\lim_{n\to\infty} \log t=\lim_{n\to\infty} (\log (n+1)-(n+1)\log (n+1)+n\log n)=-\lim_{n\to\infty} \log (1+\frac1n)^n=-1$$
so
$$\lim_{ n \to \infty} (\frac{n!}{n^n})^{\frac{1}{n}}=\frac 1e$$
another method
Lemma if $\lim\limits_{n\to\infty}\frac{a_{n+1}}{a_n}=l$, then $\lim\limits_{n\to\infty}\sqrt[n]{a_n}=l$
Let $a_n=\frac{n^n}{n!}$, it is easy to see
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to\infty}(1+\frac1n)^n=e$$
so
$$\lim_{n\to\infty}\sqrt[n]{a_n}=e$$
A: See Stirling's approximation.
A: Using inequality $$ (1+\frac{1}{n})^n < e$$  is shown by mathematical induction inequality $$(\frac{n}{e})^n <n!<e(\frac{n+1}{e})^{n+1}$$ where $n$ is a integer positive  strictly. From this inequality it follows that $$\frac{1}{e} <(\frac{n!}{n^n})^{\frac{1}{n}}<\frac{1}{e}.\frac{n+1}{n}.(n+1)^\frac{1}{n}.$$Now applying Sandwich Theorem  we obtain $$\lim (\frac{n!}{n^n})^{\frac{1}{n}} =\frac{1}{e}.$$
(It is a solution that uses the simplest means)
A: Let $$L=\lim_{n\rightarrow\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}}$$ and consider $\ln(L)$. $$\ln(L)=\lim_{n\rightarrow\infty}\frac{1}{n}\ln\left(\frac{n!}{n^n}\right)\\=\lim_{n\rightarrow\infty}\frac{1}{n}\left[\ln\left(\frac{1}{n}\right)+\ln\left(\frac{2}{n}\right)+\dots+\ln\left(\frac{n-1}{n}\right)+\ln\left(\frac{n}{n}\right)\right]\tag{1}$$ The expression $(1)$ is the Riemann sum equal to the integral $$\int_0^1\ln(x)dx=\ln(L)\tag{2}$$ The integral $(2)$ is equal to $$\left[x\ln(x)-x\right]^1_0=\lim_{x\rightarrow 0^+}-x\ln(x)-1\tag{3}$$ To evaluate the limit $(3)$, we can use l'Hôpital $$\lim_{x\rightarrow 0^+}\frac{\ln(x)}{\frac{1}{x}}=\lim_{x\rightarrow0^+}\frac{\frac{1}{x}}{-\frac{1}{x^2}}=\lim_{x\rightarrow 0^+}-x=0$$ Therefore $\ln(L)=0-1=-1$, and so it follows that $L=e^{-1}$.
