The n dimensional unit ball is that set of n-tuples <x_{1}, … x_{n}> such that Σ_{j} x_{j}^{2} ≤ 1.
We will call the content of that ball V_{n}.
The volume of the general n dimensional ball with radius r is, of course, r^{n}V_{n}.

We work backwards.
The derivative of (1−x^{2})^{n} is
−2nx(1−x^{2})^{n−1}.

The integral of x(1−x^{2})^{n} is thus
−(1−x^{2})^{n+1}/(2(n+1)).
This works for real values of n other than −1.

If we plot coordinates x_{1} and x_{2} on a plane we get the unit disk and all the points in the ball thus correspond to some point on the disk.
The other coordinates of the point in the ball that go to a point on the disk at radius r are constrained thus:
x_{3}^{2} + x_{4}^{2} + … + x_{n}^{2} ≤ 1 − r^{2}.
That set of points are in an n−2 dimensional sphere with volume V_{n−2}(1 − r^{2})^{(n−2)/2}.
We divide the disc into concentric annular rings with area 2πrdr with r running from 0 to 1.
For the content of the n dimensional ball we have the integral

V_{n} =

∫[r=0, 1] V_{n−2}(1 − r^{2})^{(n−2)/2} 2πrdr =

2πV_{n−2}∫[r=0, 1]r(1 − r^{2})^{(n−2)/2}dr =

2πV_{n−2}[1/(2((n−2)/2+1))] =

2πV_{n−2}/n.

If you believe that the unit 1D ball is the interval from −1 to 1, with content 2, then the 3D ball has 4/3 π which is what Pappus got.
We thus begin V_{1} = 2 to begin our inductive definition.
More generally we get
V_{5} = 2πV_{3}/5 = 2π(2πV_{1}/3)/5.

V_{2n+1} = (2π)^{n}/(n(n−2)… 1).

You may believe that a 0D ball has content 1, or if not that a 2D ball (disk) has content π.
Then we get

V_{6} = (2π)V_{4}/6 = (2π)^{2}V_{2}/(6∙4)
= (2π)^{3}V_{0}/(6∙4∙2)
= π^{3}/(3∙2∙1).

The odd and even dimensions unify to V_{n} = π^{n/2}/(n!) if we were to agree that n! = n((n−1)!) for fractional values of n and further that (−½)! = √π.
This may seem too cute except that

n! = ∫[x=0, ∞]x^{n}e^{−x}dx for positive integers
and ∫[x=0, ∞]x^{−½}e^{−x}dx= √π.

Further n! = Γ(n+1) and Γ(½) = √π officially.

Finally we have for the n dimensional ball of radius r the content
r^{n}V_{n} = r^{n}π^{n/2}/(n/2)!