Limit solved by definite integral (Demidovich) I was solving this limit from the Demidovich's book of exercises:
$$\lim_{n\to\infty} \frac{\sqrt[n]{\vphantom{\Large a}\, n!\,}}{n}$$
and I managed to get it to this state but then I got stuck:
$$e^{\frac{1}{n}(\log(n) + \log(n - 1) + ... + \log(1)) - \log(n)}$$
where $\log x$ is a natural logarithm of x. Can you please provide any hint?

Apply the ratio test to the sequence $\{a_n\}_{n\geq 1}$ given by:
$$ a_n = \frac{n!}{n^n}. $$
We have:
$$ \frac{a_{n+1}}{a_n} = \frac{1}{\left(1+\frac{1}{n}\right)^n} $$
hence $\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\frac{1}{e}$ implies $\lim_{n\to +\infty}\sqrt[n]{a_n}=\frac{1}{e}$, no integrals needed.

If you want to use an integral at all costs, notice that:
$$ \log(a_n) = \sum_{k=1}^{n}\log\left(\frac{k}{n}\right) $$
hence, by Riemann sums:
$$ \lim_{n\to +\infty}\frac{\log(a_n)}{n} = \int_{0}^{1}\log(x)\,dx = \color{red}{-1}.$$
