I want to fin the following integral:
$$\int_o^{x_n}\int_0^{x_{n-1}}\cdots\int_0^{x_2}n!\lambda^ne^{-\lambda(x_1+x_2+\cdots+x_n)}dx_1dx_A\cdots dx_{n-1}$$ I think that the answer is $\lambda ne^{-\lambda x}(1-e^{-\lambda x})^{n-1}$ But I didn´t find the efficient way to solve it.
Observe that by letting $f_{X}(x_i)=\lambda \mathsf e^{-\lambda x_i}$ we have: $$\begin{align} & ~~ f_{X}(x_n)~n ~(n-1)!~\iiint\limits_{0<x_1<x_2<\ldots<x_n} \prod_{i=1}^{n-1}~f_{X}(x_i)~\mathrm d x_1 \cdots \mathrm d x_{n-1} \\[1ex] = & ~ n~f_{X}(x_n)~\prod_{i=1}^{n-1}\left(\int_0^{x_n} f_{X}(x_i)\operatorname d x_i\right) \tag {$\bigstar$} \\[1ex] = & ~ n ~ f_X(x_n)~F_X(x_n)^{n-1} & : F_X(x) = \int_0^x f_X(s)\operatorname d s \\[1ex] = & ~ n\lambda ~\mathsf e^{-\lambda x_n}~ {\left(1-\mathsf e^{-\lambda x_n}\right)}^{n-1} \end{align}$$
This is the density function of the largest order statistic, $f_{\lower{0.5ex}{X_{(n)}}}\!\!(x_n)$, for a sample of $n$ independent and identically distributed exponential random variables.
$\bigstar$ You have an integral over the domain formed by $n-1$ arguments in a particular order, $0<x_1<x_2<\cdots<x_{n-1}<c$, and the function being integrated is symmetrical for those arguments $(x_1,..,x_{n-1})$.
Because of the symmetry, the integrals for domains of all orders of those arguments are identical. So $(n-1)!$ times the integral over one such domain is equal to the integral over the union of the domains. Retorically we ask: what is this union?
There are $(n-1)!$ ways these $(n-1)$ arguments could be ordered, and so the union of these $(n-1)!$ domains is $(x_1,x_2,...,x_{n-1})\in(0;c)^{n-1}$
A trivial demonstration, when $\forall x{\in}(0;c)~\forall y{\in}(0;c): Q(x,y)=Q(y,x)$, then: $$\{(x,y): 0<x<y<c\} ~\cup~\{(x,y): 0<y<x<c\} ~\equiv~\{(x,y): 0<x<c, 0<y<c\} \\ 2\int_{0<x<y<c}Q(x,y)\operatorname d (x,y) = \iint_{0<x<c,0<y<c} Q(x,y)\operatorname d (x,y)$$ And so forth for multiple arguments.
Thus $$\begin{align}& ~ (n-1)!~\iiint\limits_{0<x_1<\dots<x_{n-1}<x_n} f_X(x_1)~\cdots~f_X(x_{n-1})\mathrm d(x_1,..,x_{n-1}) \\[1ex] = & ~ \iiint\limits_{(x_1,..x_{n-1})\in(0;x_n)^{n-1}} f_X(x_1)~\cdots~f_X(x_{n-1})\mathrm d(x_1,..,x_{n-1})\\[1ex] = & ~ \left(\int_0^{x_n} f_X(x)\operatorname d x\right)^{n-1}\end{align}$$