Finding the expected value of the $\gamma$:th power of the $k$:th standard exponential order statistic in a sample of $n$ implies evaluating the integral:
$E(X_{[k]}^{\gamma})=\frac{n!}{(k-1)!(n-k)!}\int_0^{\infty}{x^{\gamma}(1-e^{-x})^{k-1}e^{-x(n-k+1)}}dx$
... does anyone have a clue for me on how to do this..? Would be much appreciated!
Note that, for $\lambda>0$, we have
$$ \int\limits_{0}^{\infty}e^{-x\lambda}\,\text{d}x{}={}\dfrac{1}{\lambda}\,. $$
So, by differentiating both sides w.r.t $\lambda$ (you can differentiate under the integral sign) "$\gamma$-times", observe that
$$ \int\limits_{0}^{\infty}x^{\gamma}e^{-x\lambda}\,\text{d}x{}={}\dfrac{\gamma!}{\lambda^{\gamma{}+{}1}}\,. $$
Since your integral can be re-written as
$$ n{n-1\choose k-1}\sum\limits_{r=0}^{k-1}\left({k-1\choose r}(-)^r\int\limits_{0}^{\infty}x^{\gamma}e^{-x(r+n-k+1)}\,\text{d}x\right)\,, $$
use the previous result to complete the integration.