Studying $ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $ I would like to find a simple equivalent of:
$$ u_{n}=\frac{1}{n!}\int_0^1 (\arcsin x)^n \mathrm dx $$
We have:
$$ 0\leq u_{n}\leq \frac{1}{n!}\left(\frac{\pi}{2}\right)^n \rightarrow0$$
So $$ u_{n} \rightarrow 0$$
Clearly:
$$ u_{n} \sim \frac{1}{n!} \int_{\sin(1)}^1 (\arcsin x)^n \mathrm dx $$
But is there a simpler equivalent for $u_{n}$?
Using integration by part:
$$ \int_0^1 (\arcsin x)^n \mathrm dx = \left(\frac{\pi}{2}\right)^n - n\int_0^1 \frac{x(\arcsin x)^{n-1}}{\sqrt{1-x^2}} \mathrm dx$$
But the relation 
$$ u_{n} \sim \frac{1}{n!} \left(\frac{\pi}{2}\right)^n$$
seems to be wrong...

This is not a complete answer, but an improved inequality. From
$$
\arcsin x\le \frac{\pi}{2}\,x
$$
you get
$$
u_n\le\frac{1}{(n+1)!}\Bigr(\frac{\pi}{2}\Bigl)^n.
$$
