$E_n =\lbrace X_n > X_m \ \forall m < n \rbrace $ are independent I'm stuck with this exercise.
Suppose $(X_n)$ are independent random variables defined on $(\Omega, \mathfrak{F}, P)$ with the same p.d.f.
Let $E_1 = \Omega$ and for $n \geq 2$ $$E_n =\lbrace X_n > X_m \ \forall m < n \rbrace $$ Then $E_1, E_2 , \ldots $ are independent with $P(E_n) = \frac{1}{n} $.
How to proceed ?
 A: The $X_{n}$ are independent and absolutely continuous. This gives
us the liberty to start with the assumption that $X_{n}\neq X_{m}$
for each pair $\left(n,m\right)$ where $n\neq m$ without any loss
of generality.
Note that the event $E_n$ can be interpreted as: $X_n$ is the "greatest" of $X_1,\dots,X_n$. 
The $X_i$ (for $i\in\{1,\dots,n\}$) all have equal chances to be the "greatest" here and one of them a.s. is the greatest. That gives $P(E_n)=\frac{1}{n}$
Every order $X_{\sigma(1)}<X_{\sigma(2)}<\cdots<X_{\sigma(n)}$ where $\sigma$ is a permutation on $\{1,\cdots,n\}$ has the same probability to occur and there are $n!$ possibilities. So denoting  $B_{\sigma}:=\{X_{\sigma(1)}<X_{\sigma(2)}<\cdots<X_{\sigma(n)}\}$ we have $P(B_{\sigma})=\frac{1}{n!}$ for each $\sigma$. Note that $\sigma\neq\sigma'\Rightarrow B_{\sigma}\cap B_{\sigma'}=\emptyset$.
It is to be shown that for each $n\in\mathbb N$ we have $P\left(A_{1}\cap\cdots\cap A_{n}\right)=P\left(A_{1}\right)\cdots P\left(A_{n}\right)$
if $A_{i}\in\left\{ E_{i},E_{i}^{c}\right\} $ for $i=1,\dots,n$.
Using induction on $n$ it is enough to prove that $P\left(A_{1}\cap\cdots\cap A_{n}\cap E_{n+1}\right)=P\left(A_{1}\cap\cdots\cap A_{n}\right)P\left(E_{n+1}\right)$
since that implies that also $P\left(A_{1}\cap\cdots\cap A_{n}\cap E_{n+1}^{c}\right)=P\left(A_{1}\cap\cdots\cap A_{n}\right)P\left(E_{n+1}^{c}\right)$. 
We have: $P\left(B_{\sigma}\cap E_{n+1}\right)=P(\left\{ X_{\sigma\left(1\right)}<\cdots<X_{\sigma\left(n\right)}<X_{n+1}\right\}) =\frac{1}{\left(n+1\right)!}=P\left(B_{\sigma}\right).\frac{1}{n+1}=P\left(B_{\sigma}\right)P\left(E_{n+1}\right)$
Now realize that every event $A_{1}\cap\cdots\cap A_{n}$ can be written as a finite union of disjoint
events $B_{\sigma}$. An immediate consequence is that $P\left(A_{1}\cap\cdots\cap A_{n}\cap E_{n+1}\right)=P\left(A_{1}\cap\cdots\cap A_{n}\right)P\left(E_{n+1}\right)$. This completes the proof.
A: From the onset, due to the $X_n$ being i.i.d. continuous random variables, $X_n{}={}X_{n'}$ for $n \ne n^{'}$ are zero probability events. Consequently, we concentrate on orderings of these random variables. 
For a given $n$, there are $n!$ exhaustive ways of ordering the random variables $X_1,\ldots,X_n$. By symmetry (from the random variables being i.i.d.) each of these orderings is identical in probability, a probability which we label $\rho_n$, and exhaustiveness implies

$$
(n!)\rho_n{}={}1\iff\rho_n{}={}\frac{1}{n!}\,.
$$

And, since by fixing $X_n$ to be the largest of these, there are $(n-1)!$ ways of arranging the order of the remaining random variables, we have
$$
P\left(E_n\right){}={}\rho_n(n-1)!{}={}\frac{1}{n!}(n-1)!{}={}\dfrac{1}{n}\,.
$$

 
Finally, choose any finite set of distinct natural numbers $n_1<\ldots<n_m$ and observe that: 
(i) There are $(n_1-1)!$ distinct ways of ordering the random variables $X_1,\ldots,X_{n_{1}-1}$, keeping $X_{n_1}$ maximum.

(ii) For each such ordering of $X_1,\ldots,X_{n_{1}-1},\,\,$ there are $\dfrac{(n_2-1)!}{n_1!}$ ways of ordering the random variables $X_1,\ldots,X_{n_{2}-1},\,\,$ keeping $X_{n_{2}}$ maximum. 
Analogous dependent counts hold for $\,n_3\,\,$ (for each "$n_2-1$" ordering), $\,n_4\,$ (for each  "$n_3-1$" ordering), and so on till $\,n_m\,$ (for each "$n_{m-1}-1$" ordering).

(iii) There are $\,\,n_m!\,\,$ exhaustive ways in which the random variables $X_1,\ldots,X_{n_m}$ may be ordered, so $\rho_{n_m}{}={}\dfrac{1}{n_m!}$.
Therefore, 

$$
\begin{eqnarray*}
P\left(E_{n_1},\ldots,E_{n_m}\right)&{}={}&\rho_{n_m}\dfrac{(n_m-1)!}{n_{m-1}!}\ldots\dfrac{(n_2-1)!}{n_1!}(n_1-1)!\newline
&&\newline
&{}={}&\dfrac{1}{n_m!}\dfrac{(n_m-1)!}{n_{m-1}!}\ldots\dfrac{(n_2-1)!}{n_1!}(n_1-1)!\newline
&&\newline
&{}={}&\dfrac{1}{n_1\ldots n_m}\newline
&&\newline
&{}={}&P\left(E_{n_1}\right)\ldots P\left(E_{n_m}\right)\,.
\end{eqnarray*}
$$

