Not asking for the distribution of $m$th order statistic, for it has been well documented and covered and in fact well-known. What I am rather confused, perhaps basic but failing to see, is the approach using the argument of symmetry in the book, An Introduction to Order Statistics:
Let $f_{m:n}$ be the pdf of the $m$th order statistic. It is given by
$$f_{m:n}(x) =n! f(x) \int\cdots\int \prod_{k=1}^{m-1} f(x_k) \prod_{k=m+1}^n f(x_k) ~\mathrm dx_1\cdots\mathrm dx_{m-1}\mathrm dx_{m+1}\cdots\mathrm dx_n, $$ over the domain $-\infty <x_1<\cdots<x_{m-1}<x<x_{m+1}<\cdots x_n<\infty.$
The authors argued that
the symmetry of $\prod_{k=1}^{m-1} f(x_k) $ with respect to $x_1, \ldots, x_{m-1}$ as well as the symmetry of $\prod_{k=m+1}^n f(x_k)$ with respect to $x_{m+1}, \cdots, x_n$ help us to evaluate the integral as follows: $$\int\cdots\int \prod_{k=1}^{m-1} f(x_k) \prod_{k=m+1}^n f(x_k) ~\mathrm dx_1\cdots\mathrm dx_{m-1}\mathrm dx_{m+1}\cdots\mathrm dx_n=\frac{1}{(m-1)! } \prod_{k=1}^{m-1} \int_{-\infty}^xf(x_k)~\mathrm dx_k\frac{1}{(n-m)!}\prod_{k=m+1}^n \int_x^\infty f(x_k)~\mathrm dx_k$$
and the form of the distribution follows.
I couldn't understand their manipulation of the integral using symmetry: what is actually the argument of symmetry suggested here? Since, they are iid, they are exchangeable but I don't know what actually the authors did. The $x_i$s are ordered in the domain: how could we permute that without not breaking the order?
Finally, how did they calculate the integral: where did the factor of ${(m-1)! }^{-1}$ and ${(n-m)! }^{-1}$ come from?
I would appreciate if someone is able to shed light on this approach.
The unrestricted integrals over $(-\infty,x)$ and $(x,\infty)$ allow all $(m-1)!$ possible orders of the first $m-1$ variables and all $(n-m)!$ possible orders of the last $n-m$ variables. Since the variables are exchangeable, the restriction to any given order yields the same contribution as the restriction originally imposed. Thus, all $(m-1)!$ contributions to the first integral are equal, and all $(n-m)!$ contributions to the second integral are equal, so in each case we can use the entire integral and divide by the number of equal contributions.