A proof involving nested integrals and induction Prove that 
$$\int_0^x dx_1 \int_0^{x_1}dx_2 \cdots \int_0^{x_{n-1}}f(x_n) \, dx_n =\frac{1}{(n-1)!}\int_0^x (x-t)^{n-1}f(t) \, dt$$
I'm trying induction over $n$. The case $n=1$ is trivial. 
When $n=2$ 
\begin{align}\int_0^x dx_1 \int_0^{x_1}f(x_{2})\,dx_2 = & \int_0^x\int_0^{x_1}f(x_2) \, dx_2 \, dx_1 \\
= & \int_0^x \int_{x_2}^{x} f(x_2) \, dx_1 \, dx_2 \\
=& \int_0^x f(x_2)(x_1-x_2) \, dx_2 = \frac{1}{(2-1)!}\int_0^x (x-t)f(t) \, dt\end{align}
I couldn't take the induction step though. Any thoughts? I appreciate the help. 
 A: The integral under consideration is
\begin{align*}
I &= \int_0^x dx_1 \int_0^{x_1}dx_2 \cdots \int_0^{x_{n-1}}f(x_n) \, dx_n & \text{} \\
&= \int_0^x \left( \ \underset{ x_n \leq x_{n-1} \leq \ldots \leq x_1 \leq x}{\idotsint} \ dx_1 \cdots dx_{n-1} \right)f(x_n) \ dx_n &  \\
&= \int_0^x \left( \ \underset{ t \leq y_1 \leq \ldots \leq y_{n-1} \leq x}{\idotsint} \ dy_1 \cdots dy_{n-1} \right)f(t) \ dt & \text{ renaming variables.}
\end{align*}
So, we will be done if we can show that
$$\underset{ t \leq y_1 \leq \ldots \leq y_{n-1} \leq x}{\idotsint} \ dy_1 \cdots dy_{n-1} = \frac{1}{(n-1)!} (x-t)^{n-1}.$$
The above integral is the volume of the $(n-1)$-simplex
$$ \{ (y_1,\ldots,y_{n-1}) : t \leq y_1 \leq \ldots y_{n-1} \leq x\} \subset [t,x]^{n-1}.$$
There is a nice trick for computing this volume geometrically, if you like that sort of thing. Since the volume of the containing cube $[t,x]^{n-1}$ is $(x-t)^{n-1}$, we might ask why the volume of the simplex is a fraction $\frac{1}{(n-1)!}$ a fraction of that? The point here is that any permutation of the variables $y_1,\ldots,y_{n-1}$ in the inequality $t \leq y_1 \leq \ldots \leq y_{n-1} \leq x$ yields another simplex with the same volume. Moreover:


*

*the volume of the overlap of any two of these simplices is zero (they can only touch on the zero volume subset of $[t,x]^{n-1}$ where at least two of the coordinates are equal) and 

*their union is the whole cube $[t,x]^{n-1}$ (any point $(y_1,\ldots,y_{n-1}) \in [t,x]^{n-1}$ must satisfy at least one of the possible chains of inequalities). 


Since there are $(n-1)!$ simplices in total, each has volume $\frac{1}{(n-1)!} (x-t)^{n-1}$, as desired. 
