How many resulting regions if we partition $\mathbb{R}^m$ with $n$ hyperplanes?

This is a generalization of this question. So in $\mathbb{R}^2$, the problem is illustrated like so:

Here, $n = 3$ lines divides $\mathbb{R}^2$ into $N_2=7$ regions. For general $n$ in the case of $\mathbb{R}^2$, the number of regions $N_2$ is $\binom{n+1}{2}+1$. But what about if we consider the case of $\mathbb{R}^m$, partitioned using $n$ hyperplanes?

Is the answer $N_m$ still $\binom{n+1}{2}+1$, or will it be a function of $m$?

Hint: Let's consider how to prove that in $\mathbb{R}^2$, the maximum number of distinct regions is $\frac { n^2 + n + 2 } { 2} $.

First, since there are finitely many lines, we can tilt any configuration such that no line is parallel to the horizontal. Now, let's consider the 'lowest' point of any region.

If the region is bound below, than such a point must exist, is unique, and is a point of intersection of 2 lines. In fact, there is a one-to-one relation between regions that are bounded below, and these points of intersections, which gives us $ { n \choose 2 } = \frac{n^2-n} {2} $ regions.

What happens if the regions are not bounded below? How many are there? Well, let's simply insert a horizontal line way down below, and count the number of unbounded regions, by associating them to these new bounded regions. Now, let's consider the 'lowest, leftmost' point of any region. If the region is bound to the left, then such a point must exist, is unique, and is appoint of intersection of the original $n$ lines with the new horizontal line. In fact, there is a one-to-one relation between regions that are unbounded below, bounded to the left, and these points of intersections, which gives us ${n \choose 1} = n$ regions.

Finally, how many regions are not bounded below, and not bounded to the left? You should convince yourself that there is ${n \choose 0} = 1 $ regions.

Hence, the total number of ways is ${n \choose 2} + {n\choose 1} + {n \choose 0}$.

Now, you can easily see how this shows that the number of regions for the hyperplane version of your problem is $$\sum_{i=0}^m {n \choose i}. $$

Denote this number as $A(m, n)$. We will prove $A(m, n) = A(m, n-1) + A(m-1, n-1)$.

Consider removing one of the hyperplanes, the maximum number is $A(m, n-1)$. Then, we add the hyperplane back. The number of regions on the hyperplane is the same as the number of newly-added regions. Since this hyperplane is $m-1$ dimensional, this maximum number is $A(m-1, n-1)$.

So, this number satisfies $A(m, n) = A(m, n-1) + A(m-1, n-1)$, and the formula can be simply derived by induction as $\sum_{i=0}^m \binom{n}{i}$.