TITLE: Proof by induction using Fubini's Theorem
QUESTION [8 upvotes]: 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?

REPLY [4 votes]: 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
$$
V(n)=V(1,n)=\frac{1}{n!}
$$