I'm trying to show that for $X_1,...,X_n$ iid with distribution function $F$ which has density $f$, the corresponding order statistics $X_{1,n}>\cdots>X_{n,n}$, the joint density of the $k$th upper order statistics is given as $$f_{X_{1,n},...,X_{k,n}}(x_1,...,x_k) = \frac{n!}{(n-k)!} F(x_k)^{n-k}\prod_{i=1}^n f(x_i),\qquad x_1>\cdots>x_k$$ So far, I have $$\begin{align*} f_{X_{1,n},...,X_{k,n}}(x_1,...,x_k) &= \int_{x_n} \cdots \int_{x_{k+1}} f_{X_{1,n},...,X_{n,n}}(x_1,...,x_n) dx_{k+1}\cdots dx_n \\ &= \int_{x_n} \cdots \int_{x_{k+1}} n! \prod_{i=1}^n f(x_i) dx_{k+1}\cdots dx_n \\ &= \frac{n!}{(n-k)!} \prod_{i=1}^k f(x_i) \int_{x_n} \cdots \int_{x_{k+1}} \prod_{i=k+1}^n f(x_i) dx_{k+1}\cdots dx_n \\ &= \frac{n!}{(n-k)!} \prod_{i=1}^k f(x_i) \prod_{i=k+1}^n \int_{x_i} f(x_i) dx_i \\ &= \frac{n!}{(n-k)!} \prod_{i=1}^k f(x_i) F(x_k)^{n-k} \end{align*}$$
But I feel like I'm missing some arguments to justify what I'm doing. I get why the joint density of the ordered sample becomes $n! \prod_{i=1}^n f(x_i)$. Can someone help me getting over the finish line?