Trouble understanding result in Simes (1986) $pr\{ P_{(j)} > \frac{j\alpha}{n} ; j = 1,...,n \} = 1-\alpha$ I'm referencing this paper http://www-stat.wharton.upenn.edu/~steele/Courses/956/ResourceDetails/MultipleComparision/Simes86pdf.pdf
On page 752, a theorem is presented which states that
If $P_{(1)},...,P_{(n)}$ are the order statistics of $n$ independent random variables from a $Unif(0,1)$ distribution, define $A_n(\alpha)$ as 
$A_n(\alpha) = pr\{ P_{(j)} > \frac{j\alpha}{n} ; j = 1,...,n \}$ for $0\leq\alpha\leq 1$
Then $A_n(\alpha) = 1 - \alpha$
The result is proved using induction, but I'm having trouble understanding all of the pieces of this proof. 
The base case (true for $n=1$) makes sense to me. I understand that $P_{(n)}$ has a cumulative distribution function of $p^n$.  
I don't understand why $\{ P_{(1)}/P_{(n)},..., P_{(n-1)}/P_{(n)} \}$ is useful or why this is a set of uniform random variables (the ratio of uniform random variables isn't uniform itself right?).
I also don't understand why we can say
$A_n(\alpha) = \int\limits_{\alpha}^1 A_{n-1}(\frac{\alpha (n-1)}{pn})np^{n-1} dp$
Even the final step (if $A_{n-1}(\alpha) = 1-\alpha$ then so does $A_n(\alpha)$) seems kind of cryptic to me.
I'd be beyond thankful if anyone could help me to fill in the blanks.
 A: I try to write some details to understand the proof. First,
$\{P_{(j)}>\frac{j\alpha}{n},j\in[1,n]\}=\{P_{(n)}>\alpha\}\cap\{\frac{P_{(j)}}{P_{(n)}}>\frac{j\alpha}{P_{(n)}n},j\in[1,n-1]\}$
Thus, $A_n(\alpha)=\mathbb{P}(\{P_{(n)}>\alpha\}\cap\{\frac{P_{(j)}}{P_{(n)}}>\frac{j\alpha}{P_{(n)}n},j\in[1,n-1]\})$
This can be rewritten as
$A_n(\alpha)=\mathbb{E}_{P_{(n)}}[\mathbb{1}_{\{P_{(n)}>\alpha\}}\mathbb{P}(\frac{P_{(j)}}{P_{(n)}}>\frac{j}{n-1}\frac{\alpha(n-1)}{P_{(n)}n},j\in[1,n-1])]$
Because, $(P_{(j)})_{j\in[1,n-1]}$ are the order statistics of n-1 independent uniform (0,1), it is the same for $(P_{(j)}/P_{(n)})_{j\in[1,n-1]}$. Indeed, if $(P_i)$ are uniform $(0,1)$, $(P_i/P_{(n)})$ are uniform $(0,1/P_{(n)})$, $P_{(n)}$ being uniform $(0,1)$. That is why 
$\mathbb{P}(\frac{P_{(j)}}{P_{(n)}}>\frac{j}{n-1}\beta,j\in[1,n-1])=A_{n-1}(\beta)$
As a result,
$A_n(\alpha)=\mathbb{E}_{P_{(n)}}[\mathbb{1}_{\{P_{(n)}>\alpha\}}A_{n-1}(\frac{\alpha(n-1)}{P_{(n)}n})]=\int_{\mathbb{R}}\mathbb{1}_{\{x>\alpha\}}A_{n-1}(\frac{\alpha(n-1)}{x n})f_{P_{(n)}}(x)dx$
what leads to the integral.
Then $A_1(\alpha)=1-\alpha$. Using the expression of $A_n(\alpha)$, it is easy to see that $A_1(\alpha)=A_2(\alpha)$, etc.
Indeed, if $A_{n-1}(\alpha)=1-\alpha$, $A_n(\alpha)=[p^n]_{\alpha}^{1}-\alpha[p^{n-1}]_{\alpha}^1$
