Is it $1^{\infty}$ form?

My solution,

Since $\mathcal{A}(n)=\sum_{k=1}^{n-1}\frac{k}{n}\sin\left(\frac{\pi k}{n}\right)=\sum_{k=1}^{n-1}\left(\frac{n-k}{n}\right)\sin\left(\frac{\pi k}{n}\right) \Rightarrow \mathcal{A}(n)=\frac{1}{2}\sum_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right)$ Now $\sum_{k=1}^{n-1}\sin\left(\frac{ \pi k}{n}\right)=\Im\left(\sum_{k=1}^{n-1} e^{\frac{i k\pi}{n}}\right)=\Im\left(\frac{1-e^{i\pi}}{1-e^{i \pi/n}}\right)=\frac{\Im\left(ie^{-i\frac{\pi}{2n}}\right)}{\sin\left(\frac{\pi}{2n}\right)}$ simplification gives $\mathcal{A}(n)=\frac{1}{2}\cot\left(\frac{\pi}{2n}\right)$ so we are now supposed to find $L=\lim_{n\to \infty}\left(\lim_{x\to \frac{\pi}{2n}}\frac{\cot (x)}{\cot\left(\frac{\pi}{2n}\right)}\right)^{\frac{1}{\tan(2nx)}}=\lim_{n\to \infty}\left(\mathcal{B}(n)\right)^{\frac{1}{\tan(2nx)}}$ We note that as $x\to \frac{\pi}{2n}\ , \mathcal{B}(n)\to 1$ then limit attains the form of $1^{\infty}$ so $\log\mathcal{B}(n)= \lim_{x\to \frac{\pi}{2n}}\frac{1}{\tan(2nx)}\log\left(\frac{\cot(x)}{\cot\left(\frac{\pi}{2n}\right)}\right)$ as $\mathcal{B}(n)$ attains $0/0$ form so due to L-hopitals rule we have $\log\mathcal{B}(n)=-\lim_{x\to\frac{\pi}{2n}}\frac{\csc(x)\sec(x)}{2n\sec^2(2nx)}= -\frac{\csc\left(\frac{\pi}{2n}\right)\sec\left(\frac{\pi}{2n}\right)}{2n}$ Therefore $L= \lim_{n\to \infty}\exp\left(-\frac{\csc\left(\frac{\pi}{2n}\right)}{2n}\right)=e^{-\pi^{-1}}=\frac{1}{\sqrt[\pi]{e}}$ is the answer.

Here the is solution due to proposer.