Division of a 3d hypercube using hyperplanes We can define a four-dimensional $hypercube$ as the set of all vectors (w, x, y, z) where each variable is restricted to 0 ≤ w, x, y, x ≤ 1.  A $hyperplane$ is the set of vectors (w, x, y, z) satisfying any linear equation, such as aw + bx + cy + dz = e.
Let $h(n)$, for any natural n, be the maximum number of pieces into which we can divide a hypercube using n hyperplanes.  
Clearly h(0) = 1, h(1) = 2, h(2) = 4, h(3) = 8, and h(4) = 16.
But what would h(6) be?
 A: The hypercube is useless; we might just as well be asking about the number of pieces into which we can divide the entire hyperspace ($\mathbb R^4$) by the said hyperplanes.
Now let's approach it the long way. Do we know into how many pieces a line can be divided by $n$ points? Surely we do: $n+1$ (some of them finite, some infinite; that's not important). OK, now let's move on to plane and divide it by lines. Once we have $n$ lines, suppose we add a new line in such a way that in intersects them all and gets cut into $n+1$ pieces. Now, each new piece of a line is cutting an old piece of plane in two, which means it increases the number of pieces of plane by 1, which means the whole line increases it by $n+1$. Can we add those $(n+1)$'s together and see how much pieces of plane we may have after $n$ linear cuts? Sure we can: that would be $n^2+n+2\over2$.
Now let's move on to our space ($\mathbb R^3$) and cut it with planes. Suppose $n$ planes cut the space into $P(n)$ pieces (areas). How many new pieces we may get by adding one more plane - in other words, can we find $P(n+1)-P(n)$? See, that new plane intersects with each of the existing ones, and each intersection is a line, so our plane has $n$ lines on it. These lines divide it into a certain number of regions, and that number is also the number of new areas of space that we got by adding that plane...
I think you got the pattern, so I don't really have to finish it for $\mathbb R^3$ and start all over for $\mathbb R^4$.
