I have a hypothesis that if: $$ I_{n,m} := n \int_{0}^{1} [1 - x^m ]^n x^m dx $$
where $m,n \in \mathbb{N}$
then $$ \lim_{n \rightarrow \infty} \frac{I_{n,m}}{n^{-\frac{1}{m} } } = c_m $$
But I have no idea how to prove it.
For $m = 1$ I can evaluate the integral using integration by parts:
$$ \int_{0}^{1} (1 - x)^nx dx = \left [- \frac{( 1-x )^{n+1} }{n+1} x \right ]_0^1 - \int_{0}^{1} -\frac{( 1-x )^{n+1} }{n+1} dx= 0 + \frac{1}{n+1} \int_{0}^{1} {(1 - x)}^{n+1} dx = \frac{1}{n+1} \left [ -\frac{( 1-x )^{n+2} }{n+2} \right ]_0^1 = \frac{1}{(n+1)(n+2)} $$
therefore:
$$ I_{n,1} = \frac{n}{(n+1)(n+2)} $$
and this can be approximated by $\frac{1}{n}$ as the hypothesis suggests.
But if I try integrating by parts for general $m$ I just transform the integral to an expression containng:
$$ \int_{0}^{1} [1 - x^m ]^n dx $$
which I do not know how to solve.
I really do not know if my hypothesis is true. I would just like to know how does $I_{n,m}$ behave for large values of $n$.
For $m > 0$, you can substitute $t = x^m$ and get
$$I_{n,m} = n \int_0^1 (1-x^m)^n x^m\,dx = \frac{n}{m} \int_0^1 (1-t)^n t^{1/m}\,dt = \frac{n}{m}B(n+1,1+1/m).$$
With the representation
$$B(p,q) = \frac{\Gamma(p)\Gamma(q)}{\Gamma(p+q)}$$
that becomes
$$I_{n,m} = \frac{n}{m}\frac{\Gamma(n+1)\Gamma(1+1/m)}{\Gamma(n+2+1/m)}.$$
Now using Stirling's formula
$$\Gamma(z+1) \sim \sqrt{2\pi} z^{z+1/2}e^{-z},$$
we obtain
$$\begin{align} n^{1/m}I_{n,m} &= \frac{\Gamma(1+1/m)}{m}\cdot \frac{n^{1+1/m}\Gamma(n+1)}{\Gamma(n+2+1/m)}\\ &\sim \frac{\Gamma(1+1/m)}{m}\cdot \frac{n^{n+3/2+1/m}e^{-n}}{(n+1+1/m)^{n+3/2+1/m}e^{-(n+1+1/m)}}\\ &= \frac{\Gamma(1+1/m)}{m}\cdot \frac{e^{1+1/m}}{\left(1 + \frac{1+1/m}{n}\right)^{n+3/2+1/m}}\\ &\to \frac{\Gamma(1+1/m)}{m}. \end{align}$$