Proof by induction using Fubini's Theorem I am asked for the volume of the region $x_1+\cdots+x_n\leq 1$ where $x_1,...,x_n\geq 0$. I am proposing that the volume $V(n)$, is given by
$$
V(n) = \int\limits_0^1\int\limits_0^{(1-x_1)}\cdots\int\limits_0^{(1-\cdots-x_{n-1})} \,dx_n\cdots\,dx_2\,dx_1 = \frac{1}{n!} \ .
$$
I am trying to prove the formula by induction. The base case is easy, but I am having a problem showing that if $n=k$ holds, then $n=k+1$ holds. I cannot figure out how to apply the inductive hypothesis. Am I missing something obvious or is there an easier method?
 A: This is example of inventor's paradox, when it is easier to prove more general fact than more specific. Let's prove by induction on $n$ that
$$
V(a,n) = \int\limits_0^a\int\limits_0^{(a-x_1)}\cdots\int\limits_0^{(a-\cdots-x_{n-1})} dx_n \cdots dx_2 dx_1 = \frac{a^n}{n!}.
$$
Basis of induction is obvious
$$
V(a,1)=\int_0^a dx_1=a
$$
Step of induction
$$
\begin{align}
V(a,n+1)&=\int\limits_0^a\int\limits_0^{(a-x_1)}\cdots\int\limits_0^{(a-\cdots-x_n)} dx_{n+1} \cdots dx_2 dx_1\\
&=\int_0^a V(a-x_1,n)dx_1\\
&=\int_0^a \frac{(a-x_1)^n}{n!}dx_1\\
&=\frac{a^{n+1}}{(n+1)!}
\end{align}
$$
In particular
$$
V(n)=V(1,n)=\frac{1}{n!}
$$
A: Try the following (you need to make this rigorous). Fix $x_{1}$. If you integrate only the $n-1$ interior integrals, you get the volume for the region when $x_{2} + \cdots x_{n} \leq 1 - x_{1}$. Heuristically, this should be 
$$\frac{1}{(n-1)!}(1-x_{1})^{n-1}$$
Where the $\frac{1}{(n-1)!}$ comes from the inductive step, and the $(1-x_{1})^{n-1}$ from the fact that you are scaling each inner integral by $(1-x_{1})$. Then finish evaluating the integral,
$$\frac{1}{(n-1)!}\int_{0}^{1} (1-x_1)^{n-1} dx_{1} = \frac{1}{(n-1)!} \cdot \frac{1}{n} = \frac{1}{n!}$$
A: In the spirit of the above proof, assume $A$ is a connected region in $\mathbb{R}^{n-1}$ of area $m(A)$, and $x$ has distance $1$ from $A$. We claim that the simplex $K$ formed by joining $x$ to $A$ in $\mathbb{R}^{n}$ has volume $m(A)/n$. 
We show this via a classical scaling argument: 
$$\int_{K}\prod dx_{i}=\int_{((1-x)A,x)}m((1-x)A)dx=\int (1-x)^{n-1}m(A)dx=m(A)\int^{1}_{0}y^{n-1}dy=\frac{m(A)}{n}$$where $y=(1-x),dy=-dx$. 
