How is the volume of a cross-polytope in $\mathbb{R}^n = \frac{2^n}{n!}$? I'm fairly new to convex geometry, and I wish to prove that the volume of a cross-polytope in $\mathbb{R}^n$ is $\frac{2^n}{n!}$.
A cross-polytope is the convex-hull of $2n$ points in $\mathbb{R}^n$, namely, $(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$. Since this is the same as the unit ball of $l_1$ norm in $\mathbb{R}^n$, we denote it by $B^n_1$.
$B^n_1$ is made up of $2^n$ pieces similar to the piece whose points are in the convex hull of $(\pm 1,0,...,0), (0,\pm 1,0,...,0),...,(0,...,0,\pm 1)$ and are all non-negative. So the total volume is $2^n$ times the volume of this piece (denoted by $P_n$ hereafter), which (from what I understand) has vertices $(1,0,...,0), (0, 1,0,...,0),...,(0,...,0, 1)$.
How do I find the volume of $P_n$? I have read an inductive proof that I don't quite understand. It says that this piece $P_n$ is a cone of height $1$ (how?) over a base, which is the analogous piece in $\mathbb{R}^{n-1}$, i.e. $P_{n-1}$ (how?).
If we take the above for granted and use that the volume of a cone in $\mathbb{R}^n$ is $\frac{Bh}{n}$ (and how do we prove this?), we'll get the desired result ($B$ = volume of base, $h$ = height of the cone) with the help of $$\operatorname{vol}(P_n) = \frac{\operatorname{vol}(P_{n-1})}{n}$$
A cone is defined as the convex hull of a single point and a convex body of dimension $\mathbb{R}^{n-1}$ in $\mathbb{R}^n$.
In general, I have found it difficult carrying forward results from $\mathbb{R}^3$ to $\mathbb{R}^n$ -  a lot of the stuff is not intuitive!
Edit 1:
Based on a comment by @Arthur, I have tried out integration to find the volume of $P_n$ as follows:
$$\int \int ... \int dx_1 dx_2 ... dx_n$$
We have $x_1 + x_2 + ... + x_n = 1$ and $x_i \ge 0 \forall i = 1,...,n$. So the limits for $x_1$ goes from $0$ to $1 - x_2 - x_3 - ... - x_n$
We get
$$\int \int ... \int (1 - x_2 - x_3 - ... - x_n) dx_2 ... dx_n$$
How do I go ahead from here? What are the limits for $x_2$? Is there a pattern I should see?
 A: These are fairly intuitive inductions if you go from $1$ to $2$ dimensions and from $2$ to $3$ dimensions.  Similar steps in higher dimensions are essentially the same.

For $n=1$ you are measuring the length from $-1$ to $+1$ which is $2=\frac{2^1}{1!}$
For the induction step you start at $n$ dimensions with a cross-polytope in black with hypervolume $V_{n} = \frac{2^n}{n!}$.  You then introduce an orthogonal  vertical line in red in the $n+1$th dimension with height $h$ from $-1$ to $+1$ and see that similar copies in grey of the cross-polytope have linear proportion $1-|h|$ and so $n$-dimensional hypervolume proportion $(1-|h|)^n$; it is easier just to consider the top half and double the result.  So integrating over the slices to find the next hypervolume: $$V_{n+1} = 2\int\limits_{h=0}^1  \frac{2^n}{n!}(1-h)^n\,dh = 2\left[\frac{2^n}{n!} \times \frac{-1}{n+1}(1-h)^{n+1}\right]^1_0=\frac{2^{n+1}}{(n+1)!}$$
A: As you've described, it's enough to find the volume of the convex hull $P_n$ of the points $(0, \ldots, 0), (1, 0, \ldots, 0), (0, 1, 0, \ldots, 0), (0, \ldots, 0, 1)$, which is sometimes called the standard $n$-simplex.

It says that this piece $P_n$ is a cone of height $1$ (how?) over a base, which is the analogous piece in $\Bbb R^{n−1}$, i.e. $P_{n−1}$ (how?).

If we think of $\Bbb R^{n - 1}$ embedded in $\Bbb R^n$ in the usual way (as the $(n - 1)$-plane $\{x_n = 0\}$, via the usual inclusion $$\iota : \Bbb R^{n - 1} \hookrightarrow \Bbb R^n, \qquad (x_1, \ldots, x_{n - 1}) \mapsto (x_1, \ldots, x_{n - 1}, 0) ,$$
then we can see that $P_n$ is the cone over the $(n - 1)$ simplex $\iota(P_{n - 1}) \subset \iota(\Bbb R^{n - 1})$ with vertex $(0, \ldots, 0, 1)$, i.e., it is a cone (in fact, pyramid) of height $1$ over $\iota(P_n)$.

If we take the above for granted and use that the volume of a cone in $\Bbb R^n$ is $\frac{B h}{n}$ (and how do we prove this?)

To prove the cone formula, suppose a cone has height $h$ and base $S$ of $(n - 1)$-dimensional volume $B$, and consider the cross-section of the cone at height $x_n$. That cross section is similar to $S$, and by linearity any of its length dimensions is $(1 - \frac{x_n}{h})$ that of the corresponding length of the base. So, the $(n - 1)$-dimensional volume of the cross-section is $\left(1 - \frac{x_n}{h}\right)^{n - 1} B$. Integrating over the $x_n$-coordinate gives that the volume of the cone is
$$\int_0^h \left(1 - \frac{x_n}{h}\right)^{n - 1} B \,dx_n = \frac{B h}{n}.$$
