If we arrange $n$ hyperplanes in a $d$-dimensional space such that the number of regions is maximized, we obtain
$$R_d(n) = \sum\limits_{i = 0}^d\binom{n}{i}$$
This is a standard result that you can find in nearly any introductory combinatorics book. This is an outline of the proof:
The following recurrence relation describes this problem:
$$R_d(n) = R_d(n-1) + R_{d-1}(n-1)$$
To see why this is true, imagine taking $n-1$ hyperplanes in general position, then adding the $n^{th}$ hyperplane to the group. If we consider only this hyperplane, each intersection with the other hyperplanes is a "hyperline" in this $(d-1)$-dimensional hyperplane, which will make $R_{d-1}(n-1)$ regions on that plane.
After you convince yourself that each of these regions created on the $n^{th}$ hyperplane adds one to the original $d$-dimensional space, the recurrence is clear. This leaves us with the following:
$$\sum\limits_{i = 0}^d\binom{n}{i} = \sum\limits_{i = 0}^d\binom{n-1}{i} + \sum\limits_{i = 0}^{d-1}\binom{n-1}{i}$$
$$= \sum\limits_{i = 0}^d\binom{n-1}{i} + \sum\limits_{i = 1}^{d}\binom{n-1}{i-1}$$
$$= \binom{n-1}{0} + \sum\limits_{i = 1}^d\binom{n-1}{i} + \binom{n-1}{i-1}$$
Apply Pascal's formula, and note that $\binom{n-1}{0} = \binom{n}{0}$
$$= \binom{n}{0} + \sum\limits_{i = 1}^d\binom{n}{i}$$
$$= \sum\limits_{i = 0}^d\binom{n}{i}$$
Because the recurrence is satisfied and the initial values match, the given summation will work (this is a very lazy way to show it holds).
For circles in the plane, the number of regions is given by A014206.
$$R(n) = n^2-n+2$$
This is easy to show by induction. The base case $n=1$, should be clear. We will assume that
$$R(n) = n^2-n+2$$
If we add another circle, then it intersects each other circle twice, creating $2$ regions (overlap, and the non-overlap area inside the new circle). As we currently have $n$ circles, adding the next will create $2n$ more regions:
$$R(n) + 2n = n^2 + n + 2 = (n+1)^2 - (n+1) + 2 = R(n+1)$$
So we are done by induction.
I don't think the statement about the tetrahedra is true. If you have just two hyperplanes, it does not hold, nor even with three. Perhaps there is further detail in your textbook?
Best Answer
Your question is equivalent to the following:
Appropriately these are called the cake numbers $C(n)$, and are indexed as A000125 in Sloane's OEIS.
The proof of this starts from two dimensions with the lazy caterer's sequence (A000124), the two-dimensional equivalent of the cake number (with a pizza). To achieve the maximum number of pieces $L(n)$ with n cuts, each new cut after the first ($L(1)=2$) must cross all previous cuts and no intersections, thereby adding n pieces at cut n. It is easy to see that the total number of pieces with n cuts is the nth triangular number plus one: $L(n)=\frac{n^2+n}2+1$.
Now go to three dimensions. After the first cut ($C(1)=2$), cut n crosses all previous cuts and no (triple) intersections; the intersections of the other planes with this latest cut form a division of the plane into the greatest number of pieces with $n-1$ cuts. So cut n adds $L(n-1)$ pieces, and after we solve the recursive equation we get $$C(n)=\frac{n^3+5n+6}6$$