I will try to solve the problem in terms of "how many regions add if we insert an extra hyperplane L (Let) in a space of k-1 hyperplanes in n-dimensions".
Let T (n,k) be the maximum number of regions formed in R^n by k hyper-planes.
Now we need to find T (n,k) - T(n,k-1)=S (Let).
Infact the answer will come out to be T (n-1,k-1).
Let us see how.
I have tried to give a purely Geometrical argument.
The number S is actually the maximum number of regions that L can pass through.[Because every region it passes through divides into 2 and hence the count for that region is 1]
Now , After drawing the new hyperplane, in your mind's eye ,consider only the intersections of L with the remaining (k-1) hyperplanes on L.
So, L which is a (n-1) dimensional Space now have a set of (k-1) intersections [ so (k-2) dimensional lines drawn on it.]
The next step is the crucial one.
4. The number of regions formed by the (k-1) lines in on L (n-1 dimensional space) is bijective to the number of regions L passes through.
So only we need to verify why 4 happens.Others are easy to infer.
Now every region formed by the (k-1) hyperplanes in R^n has a boundary of atleast 2 hyperplanes.
Now consider L ,as L passes through such a region.Observe what happens on L, at least 2 (k-2) dimesional lines occur on L.
Now to maximize the number of regions L passes through it will intersect with all the (k-1) hyperplanes.
So consider what happens when L enters one region to another.
New lines appear.
Call 2 regions adjacent in R^n if they have atleast one common hyperplane.
Now,See that L always passes through one region to its adjacent region.
As a result the intersections all give rise to adjacent regions.
And as the passes from one region to other the one- one correspondence with regions formed by the lines on L and the regions L passes through happens (due to this adjacent behaviour of the regions,L passes through)
So the maximum number of regions that the (k-1) lines on L form is actually the number T (n-1,k-1),because L is n-1 dimensional and the lines on L are its k-1 hyperplanes.
Now we need to define T (0,k) and T (k,0).
Define T (0,k)=T (k,0)=1.(quite natural)
This seems like a kind of a hand waving argument because we cant actaully visualize n>4.
But this can be easily checked in 3 dimensions and 2 dimensions.
Note: I have mentioned lines and planes to let you visulaize in 3 dimesions.
Now in accordance with the previous answer we know that there exists a similar recursion in the combinatorial domain.
If we solve the recursion,it comes out to be the same as above.
Moreover you can substitute and check that it holds.
Will like to hear how to verify 4 in other different ways in the comments!
Will like to hear any good suggestions to this answer!
I am new to this ,If you want to downvote my answer (please mention the reason,otherwise I will repeat the same mistakes again).