Let $X_1,...,X_n\sim U[0,1]$
We have to show that:
- $$\mathbf{E}[X_{(k)}^r] =\frac{n!\Gamma(r+k)\ }{(r-1)!\Gamma(n+k+1)} $$
2.If $r$ is integer then: $$\mathbf{E}[X_{(k)}^r] = \frac{ {r+k-1 \choose r}}{ {n+r \choose r}}$$
My proof for 1.
$$\mathbf{E}[X_{(k)}^r] = \int_0^1 x^rf(x) \,dx= \frac{n!}{(k-1)!(n-k)!} \int_0^1 x^r[F(x)]^{k-1}[1-F(x)]^{n-k}f(x)dx $$ $$ = \frac{n!}{(k-1)!(n-k)!}\int_0^1 x^rx^{k-1}(1-x)^{n-k}dx $$
$$ =\frac{n!}{(k-1)!(n-k)!}\beta(k+r,n-k+1)=\frac{n!\Gamma(r+k)\Gamma(n-k+1)}{(k-1)!(n-k)!\Gamma(n+k+1)}=\frac{n!\Gamma(r+k)\ (n-k)(n-k-1)!}{(r-1)!(n-k)!\Gamma(n+k+1)} = \frac{n!\Gamma(r+k)\ }{(k-1)!\Gamma(n+r+1)}$$
but I got stuck to prove 2, can anyone help me?
You cannot get Equation $2$ from Equation $1$, because they are not the same. You've mixed up your $r$ and $k$. Notice how in your proof of Equation $1$, you wrote
$$\operatorname{E}[X_{(k)}^r] = \int_{x=0}^1 x^{\color{red}{k}} f_{X_{(k)}}(x) \, dx.$$ This is clearly wrong, because you wrote an exponent of $r$ on the left hand side, but the exponent on the right hand side is $k$. If you do the computation correctly with $r$ instead, then you will get
$$\operatorname{E}[X_{(k)}^r] = \frac{n! \, \Gamma(r+k)}{(\color{blue}{k}-1)! \, \Gamma(n+\color{blue}{r}+1)}.$$ Then this equation is consistent with Equation $2$ from your original question.