where $S_n$ is the hyper sphere area and $R$ its radius.
it's written on 'mathworld.wolfram' like it's a definition.
computing by hand the volume for $n= 1,2,3$ I can see that it holds, so is it just a generalization deduced from the observation of a pattern ? or is there some other kind of justification ?
Not everything on Wolfram Mathworld is explained in the form of a sequence of implications. My recollection is that some formulas are not proved at all.
If you "stretch" an $n$-dimensional "solid" object (such as the $n$-ball, which is a name for the interior of a hypersphere embedded in $n$-dimensional Euclidean space) by a factor of $R$ in one dimension, you get an object that has exactly $R$ times as much content (which I'll call "volume" for analogy with the $3$-ball). Now stretch the resulting object by a factor of $R$ in a second dimension, and its volume increases by a factor of $R$ again, to $R^2$ times the original volume.
But there are altogether $n$ separate dimensions in which we can stretch an $n$-ball in $n$-dimensional Euclidean space. After stretching in each of those dimensions, we have another $n$-ball that has exactly $R$ times the radius of the original ball, and $R^n$ times the volume.
In this way we can relate the volume of an $n$-ball of any radius to the volume $B_n$ of an $n$-ball of radius $1.$ The volume of the $n$-ball of radius $R$ is $$ B_n R^n. $$
Now take two concentric $n$-spherical shells within the $n$-ball at radii $r$ and $r + \Delta r.$ For small $\Delta r,$ the volume of the region between the shells is approximately $$ A(r) \Delta r,$$ where $A(r)$ is the "area" of the $n$-sphere of radius $r.$ We can confirm this by dividing the $n$-sphere into very small pieces so that the region between the shells is filled by putting an almost-prism shape on each subdivided region of the inner shell.
We can therefore integrate the volume of the $n$-ball by a method of spherical shells, which tells us that $$ B_n R^n = \int_0^R A(r) \, dr, \tag1 $$ and by the fundamental theorem of calculus, it follows that $$ A(r) = \frac{d}{dr} B_n r^n = n B_n r^{n-1}.$$
So we set $S_n = n B_n,$ interpreted as the "area" of the unit sphere in $n$-dimensional space, and then the area of the sphere of radius $r$ is $A(r) = S_n r^{n-1}.$ Make these substitutions for $B_n$ and $A(r)$ in Equation $(1),$ and you get $$ \frac{S_n}{n} R^n = \int_0^R S_n r^{n-1} \, dr.$$
The author of the Wolfram Alpha page apparently considered this obvious enough to state without proof; after that there is a derivation of a formula for $S_n.$
This answer gives a similar argument and generalizes it to polytopes (including hypercubes) as well as spheres.