I've already found out that this topic has been discussed here: What am I doing wrong finding the $n$-th derivative of $\arcsin(x)$ and its value at $0$?
However, I cannot totally understand what's going on there... It would be greatly appreciated if somebody showed it with its clear proof.
If you find the Taylor series at $0$ for the arcsine, you get $\arcsin x=\sum_{k=0}^\infty a_nx^n$, and then $\arcsin^{(n)}(0)=n!a_n$. Since $$ a_n=\frac{1 }{2^{2n}}\binom{2n}{n} \frac{ 1}{2n+1}, $$ you get that $$ \arcsin^{(n)}(0)=\frac{n! }{2^{2n}}\binom{2n}{n} \frac{ 1}{2n+1}=\frac{(2n)!}{4^n\,n!\,(2n+1).} $$
