Let $X_k$ be $\text{i.i.d.}$ continuous random variables. Find in terms of $n:$ $$\mathbb{P}\Big(X_1\geq X_2\geq\cdots \geq X_{n-1}<X_n\Big)$$
Let each $X_k$ have $\text{p.d.f }\;f$ and $\text{c.d.f}\;F$, so that: $$f_{X_1,\;\dots,\; X_n}(x_1,\;\dots,\; x_n)=f(x_1)\cdots f(x_n)$$ The probability should $(?)$ resemble something like: $$\int_{-\infty}^{\infty}\int_{-\infty}^{x_1}\cdots\int_{-\infty}^{x_{n-2}}\color{red}{\int_{?}^?}f(x_1)\cdots f(x_n)\;\color{red}{\mathrm{d}x_n}\mathrm{d}x_{n-1}\cdots\mathrm{d}x_1$$ The problem is that I cannot see how to incorporate $\color{red}{X_{n-1}<X_n}$ into the limits. Ignoring this the expression becomes: $$\int_{-\infty}^{\infty} \left(f(x_1)\mathrm{d}x_1\int_{-\infty}^{x_1}\left(f(x_2)\mathrm{d}x_2\cdots\int_{-\infty}^{x_{n-2}}f(x_{n-1})\mathrm{d}x_{n-1}\right)\cdots\right)=\color{blue}{\frac{1}{n!}}$$ By using $\int_{-\infty}^{x_{k}}f(x_{k+1})\mathrm{d}x_{k+1}=F(x_{k})$ repeatedly. I believe this is the $\color{blue}{\text{correct}}$ result, but I have arrived at it via a flawed reasoning. How can I adjust my working?
Note: I am not interested in an different approach at the minute, however slick; I'd like to see how this is done using the method above so that I gain a greater understanding of the subject.
Set the limits so that the innermost integral is over the interval $[x_{n-1},\infty)$. Then this integral is
$$\int_{x_{n-1}}^{\infty}f(x_n)dx_n = 1-F(x_{n-1})$$
The iterated integral over the remaining variables splits into two pieces resulting in
$$P= \frac{1}{(n-1)!}-\frac{1}{n!}= \frac{1}{(n-1)!}(1-\frac{1}{n})$$