Proof for convert $\left\{\int_a^tf(x)dx\right\}^{n}$ to multiplication of integrals In a book, author use a relation like this:
$$\left\{\int_a^tf(x)dx\right\}^{n} = n!\int_a^tf(x_1)dx_1\int_a^{x_1}f(x_2)dx_2\dots\int_a^{x_{n-1}}f(x)dx_{n}.$$
How I can make a proof for this? Which condition on $f$ is necessary for that equation became true?
Thanks,
Meysam.
 A: Hint: In the left-hand side, $x_1$, $x_2$,... $x_n$ can take arbitrary values between $a$ and $t$, in the right-hand side, you have $t\geq x_1\geq x_2\geq\dots\geq x_n\geq a$... and a factor $n!$ which is the number or permutations of $n$ elements.
Edit: you should first write
$$\left(\int_a^tf(x)\,\mathrm{d}x\right)^n=\left(\int_a^tf(x_1)\,\mathrm{d}x_1\right)\cdots\left(\int_a^tf(x_n)\,\mathrm{d}x_n\right),$$
which rewrites
$$\left(\int_a^tf(x)\,\mathrm{d}x\right)^n=\int_a^t\mathrm{d}x_1\dots\int_a^t\mathrm{d}x_n\;\;f(x_1)\dots f(x_n).$$
The variables $x_1$, ..., $x_n$ in the integral are in arbitary order. Reorder them in decreasing order ($t\geq x_{\sigma(1)}\geq x_{\sigma(2)}\geq\dots\geq x_{\sigma(n)}\geq a$) where $\sigma$ is a permutation of the $n$ integers $1$,... $n$. The number of such permutations is $n!$. With these elements, you should be able to conclude easily.
A: The only condition we need is $f$ Lebesgue-integrable. 
I assume without loss of generality that $a=0$ and $t=1$.
Define $g(x_1,\dots,x_n):=f(x_1)\dots f(x_n)$, $I:=[0,1]^n$ and $S:=\{(t_1,\dots,t_n)\in I,0\lt t_1\lt\dots\lt t_n\lt 1\}$. 
Then $$I':=\{(t_i)_{i=1}^n,i\neq j\Rightarrow t_i\lt t_j\}=\bigsqcup_{\sigma\in\mathcal S_n}\{(t_1,\dots,t_n),(t_{\sigma(i)})_{i=1}^n\in S\}.$$ 
Using the symmetry of $g$ and the fact that $I\setminus I'$ has Lebesgue measure $0$ we can conclude. 
