I'm trying to evaluate $L=\lim\limits_{n\to\infty}f(n)$ where
$$f(n)=\prod\limits_{k=1}^n \frac{2n}{e}\left(\arcsin\left(\frac{k}{n}\right)-\arcsin\left(\frac{k-1}{n}\right)\right)$$
We have:
$f(1)\approx1.15573$
$f(10^2)\approx1.22410$
$f(10^4)\approx1.22533$
$f(10^6)\approx1.22535$
My attempt
I've only been able to find the following equivalent expressions for $L$.
$$L=\lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n\log\left(\frac{2n}{e}\left(\arcsin\left(\frac{k}{n}\right)-\arcsin\left(\frac{k-1}{n}\right)\right)\right)\right)$$
$$L=\lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n\log\left(\frac{2n}{e}\arcsin\left(\frac{k\sqrt{n^2-(k-1)^2}-(k-1)\sqrt{n^2-k^2}}{n^2}\right)\right)\right)$$
$$L=\lim\limits_{n\to\infty}\exp\left(\sum\limits_{k=1}^n\log\left(\frac{2n}{e}\arccos\left(\frac{\sqrt{(n^2-k^2)(n^2-(k-1)^2)}+k(k-1)}{n^2}\right)\right)\right)$$
Context
The limit comes from the following geometrical construction that I made up.
In a semicircle with diameter $na$ where $a=\frac{2}{e}$, draw chords of lengths $a,2a,3a,\dots,na$ that meet at one of the endpoints of the semicircle. Here is an example with $n=6$.
Consider the arcs between neighboring chord endpoints. The product of the $n$ arc lengths is $f(n)$ from above.
Why I think $L$ might have a closed form
I think $L$ might have a closed form, because a similar geometrical construction seems to yield a limit with a closed form.
Instead of starting with a semicircle (which is a cicular segment with central angle $\pi$), let's start with a circular segment with central angle $\pi/3$, with a bounding chord of length $nb$ where $b=\frac{3\sqrt3}{2e}$. Draw chords of lengths $b,2b,3b,\dots,nb$ that meet at one of the end points of the arc. Here is an example with $n=6$.
Consider the arcs between neighboring chord endpoints. The product of the $n$ arc lengths is $g(n)=\prod\limits_{k=1}^n\frac{3\sqrt3n}{e}\left(\arcsin\left(\frac{k}{2n}\right)-\arcsin\left(\frac{k-1}{2n}\right)\right)$.
Numerical investigation suggests that $\lim\limits_{n\to\infty}g(n)=1$.
Best Answer
Long comment. What I proved in this posting about an asymptotic formula for a related product actually works for $\alpha, \beta > -1$ (see the statement of the corollary in the link). The upshot of this asymptotic formula in OP's case is that
\begin{align*} &\prod_{k=1}^{n} \frac{2n}{e} \left[ \arcsin\left(\frac{k}{n}\right) - \arcsin\left(\frac{k-1}{n}\right) \right] \\ &\sim \frac{1}{(2\pi n)^{1/4}} \prod_{k=1}^{n} \frac{2}{1 + \sqrt{1-k^{-1}}} \\ &\sim \sqrt[4]{\frac{e^{\gamma}}{2\pi}} \exp\Biggl[ \sum_{k=1}^{\infty} \log\Biggl( \frac{2}{1 + \sqrt{1-k^{-1}}} \Biggr) -\frac{1}{4k} \Biggr] \\ &\approx 1.2253517701\ldots \end{align*}
I am skeptical that this expression has an elementary closed form.