I'm trying to calculate the expectation of $k$th order statistic of exponential dist $E_k$,where the last index is $n$.(I'll fix the parameter $\theta$ as 1, since it is not big issue.)
Let $X \sim exp(1) , X_i \sim exp(1), Y = \sum_{i=1}^n I(X_i \le x)$ (I is indicator function)
then $Y \sim Binominal(n,F_{X}(x))$ since each $I(X_i \le x)$ is Bernoulli trial with probability $F_{X}(x)$.
Then $1 - F_{X_{(k)}}(x)$ can expressed as
$\sum_{i = 0}^{k-1} P(Y=i)$ since $P(X_(k) >x)$ = $P$(no $X_i$ is less then x)+ $P$(one $X_i$ is less then x)+...+$P$($(k-1) X_i$ is less then x)
By the expectation fomula of positive random variable, $E_k-E_{k-1}$ = $\int_{0}^\infty 1-F_{X_(k)}-(1-F_{X_(k-1)})dx = \int_{0}^\infty P(Y = k-1)dx$ $= \int_{0}^\infty {n \choose k-1}(F_X(x))^{k-1}(1-F_X(x))^{n-k+1}dx =\int_{0}^\infty {n \choose k-1}(1-e^{-x})^{k-1}(e^{-x})^{n-k+1}dx = \int_{0}^\infty {n \choose k-1}(1-u^{-1})^{k-1}(u)^{-(n-k+2)}du$ $= {n \choose k-1}[\frac{{{k-1} \choose 0}}{n-({k-1})}-\frac{{{k-1} \choose 1}}{n-(k-1)+1}+\frac{{{k-1} \choose 2}}{n-(k+1)+2}-...+(-1)^{k-1}\frac{{{k-1} \choose {k-1}}}{n}] $
Here's my question: I guess that the final term would be simplify to $\frac{1}{n-(k-1)}$, then it makes sence with $E_1 = \frac{1}{n}$ and $E_n = H_n$(harmonic series). but I cannot prove that $[\frac{{{k-1} \choose 0}}{n-({k-1})}-\frac{{{k-1} \choose 1}}{n-(k-1)+1}+\frac{{{k-1} \choose 2}}{n-(k+1)+2}-...+(-1)^{k-1}\frac{{{k-1} \choose {k-1}}}{n}]$ is same with $\frac{1}{(n-k){n \choose k-1}}$. How can I prove this?
