How to get value of $\lim\limits_{n \to \infty} \frac{\sqrt[n]{n!}}{n}$ without using Stirling-formula? The problem $$\lim\limits_{n \to \infty}\frac{\sqrt[n]{n!}}{n}$$ was a for-fun-only exercise given by our Calculus Professor. I was able to solve it quite easily with the use of the Stirling-formula, but can't figure out if it can be done in a different way, possibly by brute force spiced with recognizing the Euler-number in the proccess. I got to $$\lim\limits_{n \to \infty} \sqrt[n]{1 \cdot (1-1/n) \cdot (1-2/n) \cdot ... \cdot (1-(n-1)/n) }$$ by simply bringing the $n$ inside the n-th root sign, but I believe this is not the correct path.
I have no other ideas.
 A: Let $a_n = \frac{n!}{n^n}$, then 
$$\require{cancel}\frac{a_{n+1}}{a_n} = \frac{(n+1)!}{(n+1)^{n+1}} \frac{n^n}{n!} = \frac{\cancel{(n+1)}}{\frac{(n+1)^n}{n^n} \cancel{(n+1)}} = \frac{1}{(1 + \frac{1}{n})^n} $$
Now $$\lim_{n\to \infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}$$
Then $$\lim_{n\to\infty} \sqrt[n]{a_n} = \color{#05f}{\frac{1}{e}}$$
A: If $a_n = \sqrt[n]{n!}/n$, then 
\begin{align}\lim_{n\to \infty} \log a_n &= \lim_{n\to \infty} \left(\frac{1}{n}\log n! - \log n\right)\\
&= \lim_{n\to \infty} \left(\frac{1}{n}\sum_{k = 1}^n \log k - \frac{1}{n}\sum_{k = 1}^n \log n\right)\\
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k = 1}^n (\log k - \log n)\\
&= \lim_{n\to \infty} \frac{1}{n}\sum_{k = 1}^n \log \frac{k}{n}\\
&= \int_0^1 \log x\, dx\\
&= (x\log x - x)\bigg|_{x = 0}^1\\
&= -1.
\end{align}
Therefore, by continuity of $x \mapsto e^x$, 
$$\lim_{n\to \infty} a_n = \frac{1}{e}.$$
A: You could take the log of this expression, and write it as a Riemann sum :
$$\ln\left( \sqrt[n]{ \prod_{k=0}^{n-1} 1-\frac{k}{n} } \right) = \frac{1}{n} \sum_{k=0}^{n-1} \ln(1-\frac{k}{n} )$$
Hence
$$\lim_{n\to +\infty}\ln\left( \sqrt[n]{ \prod_{k=0}^{n-1} 1-\frac{k}{n} } \right) = \int_{0}^1 \ln(1-x) dx$$
A: Start as
$$ \frac{(n!)^{1/n}}{n} = e^{ \frac{1}{n}\ln n! - \ln n } \sim e^{ \frac{n\ln n-n+1}{n}-\ln n }= e^{-1+1/n} \longrightarrow ...  $$
