An interesting limit: $\lim_{n\rightarrow\infty} \left(\frac{n^{n-1}}{(n-1)!}\right)^{\frac{1}{n}}$ It was a new contributor's question. I answered, got my -1 again and then deleted. Then I asked myself. Then gave it up again. Actually I was gonna ask a different question NOW. When I pressed ask a question, to my surprise, the question I intended to ask yesterday was in the memory!
I wanted to evaluate the following limit by logarithmic limit rule:
$$\lim_{n\rightarrow\infty} \left(\frac{n^{n-1}}{(n-1)!}\right)^{\frac{1}{n}}=\exp\left(\lim_{n\rightarrow\infty}\frac{(n-1)\ln n-\ln (n-1)!}{n}\right)=\exp\left(\lim_{n\rightarrow\infty}-\frac{1}{n}\sum_{k=1}^n\ln(\frac{k}{n})\right)$$
Then I observed a Riemann sum of an indefinite integral inside so that the limit is
$$\exp\left(-\int_0^1\ln xdx\right)=\exp\left((x-x\ln x\vert_0^1)\right)=e.$$
Is my solution correct?  Can you suggest another way? Stirling's approximation formula is excluded.
 A: It is correct up to two things.
First, you should not write $\lim_{n \to +\infty}$ before the existence of the limit is proved. You have to start with
$$\Big(\frac{n^{n-1}}{(n-1)!}\Big)^{1/n} = \Big(\frac{n^{n}}{n!}\Big)^{1/n} = \exp\Big(n \ln(n) - \frac{1}{n} \sum_{k=1}^n \ln(k)\Big) = \exp\Big(- \frac{1}{n}\sum_{k=1}^n \ln(k/n)\Big).$$
Second, the theorem form the convergence of Riemann sums applies to continuous functions on $[0,1]$. Here, the function $\ln$ is continuous on $]0,1]$ only. Yet, it is increasing, so you can add inequalities
$$\int_{(k-1)/n}^{k/n} \ln x dx \le \frac{1}{n} \ln(k/n) \le \int_{k/n}^{(k+1)/n} \ln x dx.$$
over all $k \in \{1,\ldots,n\}$, to get a lower bound and an upper bound and prove the existence of the limit.
Other method :
$$\frac{(k+1)^{k+1}/(k+1)!}{k^k/k!} = \frac{(k+1)^k}{k^k} = (1+1/k)^k \to e \text{ as  } k \to +\infty,$$
so an application of Ces`ar'o lemma (take logarithms) yields
$$\Big(\frac{n^{n}}{n!}\Big)^{1/n} \to e \text{ as  } k \to +\infty.$$
A: By the Stirling formula:
$$\dfrac{n^{n-1}}{(n - 1)!} = \dfrac{n^n}{n!} \sim \dfrac{e^n}{\sqrt{2 \pi n}}$$
then :
$$\left(\dfrac{n^{n-1}}{(n - 1)!}\right)^{\frac{1}{n}} \sim \dfrac{e}{(2 \pi n)^{1/2n}} \to e$$
