a question about how to prove mutivariable integral, I am struggling about it! If $f(x)$ is Riemann integrable in $[a,b]$, and then how to prove $$\int_{a}^{b} f(x_1) \, dx_1 \int_{a}^{x_1}f(x_2) \, dx_2 \cdots \int_{a}^{x_{n-1}}f(x_n) \, dx_n={1\over n!} \left[\int_a^b f(x) \, dx \right]^n$$
I really don't know how to solve it! I prefer to use mathematical induction,but it doesn't work. Can someone can help me to solve it? I will appreciate you very much.
 A: Here is a tedious calculation:
Let $I=[a,b]$, and
Let $S = \{ x \in I^n \mid x_1 < \cdots < x_n \}$.
In the following, let $\sigma$ denote a permutation of $\{1,\ldots,n\}$.
Let $S_\sigma = \{ x \in I^n \mid (x_{\sigma_1},\ldots,x_{\sigma_n}) \in S \}$, and note that the $S_\sigma$ are disjoint.
It is not hard to see that the function $x_n \mapsto 1_{S_\sigma}((x_1,\ldots,x_n))$ is Riemann integrable.
Let $\phi(x) = f(x_1)\cdot \cdot \cdot f(x_n)$ for $x \in I^n$, and note that  $\phi((x_{\sigma_1},\ldots,x_{\sigma_n})) = \phi(x)$.
Note that $\int_a^b \int_a^{x_1} \cdots \int_a^{x_{n-1}} \phi(x) \, dx_1 \cdots d x_n = \int_a^b \cdots \int_a^b \phi(x) 1_S(x)\,dx_1 \cdots d x_n $, and that
$\int_a^b \cdots \int_a^b \phi(x) 1_S(x)\,dx_1 \cdots d x_n = \int_a^b \cdots \int_a^b \phi(x) 1_{S_\sigma}(x)\,dx_1 \cdots d x_n $ for all permutations $\sigma$.
Hence $\int_a^b \cdots \int_a^b \phi(x) 1_S(x) \, dx_1 \cdots d x_n = {1 \over n!} \int_a^b \cdots \int_a^b \phi(x) \, dx_1 \cdots d x_n $.
Since $(\int_a^b f(t) \, dt )^n = \int_a^b \cdots \int_a^b \phi(x) \, dx_1 \cdots d x_n$, we see that
$\int_a^b \int_a^{x_1} \cdots \int_a^{x_{n-1}} \phi(x) \, dx_1 \cdots d x_n
= {1 \over n!}(\int_a^b f(t) \, dt )^n$.
Note: I haven't shown that the various integrals exist, however it is sufficient to note that if $g,h$ are Riemann integrable, then $g \cdot h$ is Riemann integrable. Furthermore, if $g$ is Riemann integrable, the function $t \mapsto \int_a^t g(x) dx$ is continuous, hence Riemann integrable.
Note: I have inadvertently used Fubini above. To justify this I need a
result which is that if $(x,y) \mapsto f(x,y)$ is integrable on $A\times B$, and $x \mapsto \int_B f(x,y)dy$, $y \mapsto \int_A f(x,y)dy$ are integrable on $A$, $B$ respectively, then the iterated integrals can be swapped. In addition, I would need to use the fact that any permutation can be written as the composition of a finite number of pairwise swaps. (This is only needed with the Riemann integral, of course.)
A: With method of induction (beginning with a partial integration) :

