Let $X_1,\dots,X_n$ be continuous random variables with distribution $F$. Let $U_i=F(X_i)$. I have to show that $$ f_{(k)}(x)=\dfrac{n!}{(k-1)!(n-k)!}x^{k-1}(1-x)^{n-k}, $$ where $f_{(k)}$ stands for the PDF of the $k$th order statistic. I’ve already shown that $$ F_{(k)}(x)=\sum_{j=k}^n\binom{n}{j}x^j(1-x)^{n-j}. $$ Now I tried $f_{(k)}(x)=\dfrac{d}{dx}F_{(k)}(x)$, but I'm not sure how to differentiate the binomial. Is it possible to work with induction here? Even then I'm slightly confused, because I would say that we can only go as far up to $n$, and not $\infty$; so would induction be the right word?
Any hint is greatly appreciated.
Differentiate your sum with respect to $x$ and you'll get $$ \sum_{j=k}^n{n\choose j}\left[ jx^{j-1}(1-x)^{n-j}-(n-j)x^j(1-x)^{n-j-1}\right]. $$ Break this into the difference of two sums, the first one being $$ A:=\sum_{j=k}^n{n\choose j} jx^{j-1}(1-x)^{n-j} =\sum_{j=k}^n\frac{n!}{(j-1)!(n-j)!} x^{j-1}(1-x)^{n-j} $$ and the second one $$ B:=\sum_{j=k}^n{n\choose j}(n-j)x^j(1-x)^{n-j-1} \stackrel{(*)}=\sum_{j=k}^{n-1}\frac{n!}{j!(n-j-1)!}x^j(1-x)^{n-j-1} $$ In the sum on the left of (*) we notice the $j=n$ term is zero, so stop the sum at $j=n-1$. Next, change index of summation to $i:=j+1$ and you get $$ B=\sum_{i=k+1}^n\frac{n!}{(i-1)!(n-i)!}x^{i-1}(1-x)^{n-i}. $$ Now we see that $A$ minus $B$ is a single term, and you're done.