For the question stated in the title, the answer is yes, if more is interpreted as "more than or equal to".
Proof: let $\Lambda$ be a collection of lines, and let $P$ be the extended two plane (the Riemann sphere). Let $P_1$ be a connected component of $P\setminus \Lambda$. Let $C$ be a small circle entirely contained in $P_1$. Let $\Phi$ be the conformal inversion of $P$ about $C$. Then by elementary properties of conformal inversion, $\Phi(\Lambda)$ is now a collection of circles in $P$. The number of connected components of $P\setminus \Phi(\Lambda)$ is the same as the number of connected components of $P\setminus \Lambda$ since $\Phi$ is continuous. So this shows that for any collection of lines, one can find a collection of circles that divides the plane into at least the same number of regions.
Remark: via the conformal inversion, all the circles in $\Phi(\Lambda)$ thus constructed pass through the center of the circle $C$. One can imagine that by perturbing one of the circles somewhat to reduce concurrency, one can increase the number of regions.
Another way to think about it is that lines can be approximated by really, really large circles. So starting with a configuration of lines, you can replace the lines with really really large circles. Then in the finite region "close" to where all the intersections are, the number of regions formed is already the same as that coming from lines. But when the circles "curve back", additional intersections can happen and that can only introduce "new" regions.
Lastly, yes, the number you derived is correct. See also this OEIS entry.
With four lines, there are seven different ways, i.e. numbers of regions into which a bounded space can be divided by four lines. The minimum number of regions is $4+1=5$, produced when no lines intersect within the space.
But every number of regions from $5$ to maximum $11$ is possible. For beginning with $4$ non-intersecting lines within the space, we can add $1, 2, 3$ regions by redrawing one line so as to cut first one, then two, then three other lines, as in the top row of the figure.
Then we add a $4th$ and $5th$ region by redrawing the third line so as to cut one and then both of the first two lines, as in the second row of the figure.
Finally, we get a $6th$ region by making the second line cut the first, as in the third row.
Thus to the original five regions we have added$$3+2+1=6$$Generalizing, since the minimum number of regions is $n+1$, and the maximum is$$\frac{n^2+n+2}{2}$$then the number of ways will be$$\frac{n^2+n+2}{2}-n=\frac{n^2-n+2}{2}$$
And since$$\frac{n^2-n+2}{2}=\frac{(n-1)n}{2}+1$$we see that the number of ways $n$ lines can divide the space is equal to the $(n-1)th$ triangle number plus $1$.
Note: This seems to be something more general than a geometric problem. If the space is bounded, no pair of non-intersecting lines need to be parallel. Nor do the line segments need to be straight, provided that no two intersect more than once within the space. And perhaps the only condition on the bounded space is that it be concave from within?
Correction: If the plane is unbounded, and "how many ways" means "how many arrangements," as arrangement is explained in the comments on OEIS A241600 referenced by @Daniel Mathias, then the above is not a suitable answer to the question posted, and there do appear to be nine arrangements of four lines.
The first two rows have parallels, the third does not. There is one 3-line concurrence in the second row, and a 3-line and 4-line concurrence in the third. The number of regions, left to right and top to bottom, is$$5, 8, 9, 9, 10, 10, 8, 10, 11$$Unlike the situation as I first understood it, there are gaps and repetitions in the number of regions produced by the different arrangements. This seems to make the determination of $P$ as a function of $n$ a more difficult task.
Correction continued: $P$ for $n=5$
The figures shows twenty-one arrangements for two or more lines parallel when $n=5$. In each row (except the last, which is actually two rows of two each) four lines keep a given position throughout the row as a fifth line shifts its position through the essentially different arrangements possible. After the first row showing one arrangement each for five and four parallels, the second row has only three parallels, the third has two pairs of two, and the fourth and fifth only two.
Next we can see by a single figure below the arrangements possible when no lines are parallel. Again we suppose the four lines $AB$, $AC$, $FB$, $FD$ given in position. $G$ is any point not collinear with any two of the six intersection points $A$, $B$, $C$, $D$, $E$, $F$. A fifth line through $G$ can pass through and among the six points in $6+7=13$ different ways.
But if $G$ is collinear with a line $BE$, $DC$, or $AF$, it can be seen even without another figure that the number of possible arrangements will be only $5+6=11$.
And finally, we have one arrangement each for four and five concurrent lines.
Adding them up,$$1+1+5+4+10+13+11+1+1=47$$in agreement with OEIS A241600.
Best Answer
The general idea is as follows: You start with an original configuration of some elements (an element being a line or a circle) already in the plane, then add a new element (line or circle) to get a new configuration. The existing regions in the original configuration subdivide the new element into certain sections.
The important part is that these sections on the new element determine by how much the number of regions increases. Because each section subdivides a region from the original configurations into 2 regions of the new configuration, the number of regions increases by exactly the amount of sections that the original regions create from the new element.
OTOH, the number of those sections is easily determined by the number of common points the new element has with the elements of the original configuration.
If the new element is a line that has $l$ common points with the original configuration, then the new line will be divided into $l+1$ sections.
If the new element is a circle that has $c$ common points with the original configuration, then the new circle will be divided into $c$ sections if $c > 0$, or 1 section if $c=0$.
Let's call $R(m,n)$ the maximum amount of regions that $m$ circles and $n$ lines can divide the plane into.
We start with $R(0,0)=1$: The whole plane is one region. $R(1,0)=2$, because a single circle has an outer and an inner region.
For $m \ge 2$ we get $R(m,0) \le R(m-1,0) + 2(m-1)$. That's because the $m-1$ existing circles can have at most 2 common points with the new circle each (and $2(m-1) \ge 1$ in case the new circle has no point in common with existing circles).
This leads, per induction, to $R(m,0) \le m(m-1)+2$ for $m \ge 1$, a formula that (with equality) is also in the linked treaty. In order to have 'common' estimation also for $m=0$, let's define $s(m)=1$ for $m=0$ and $s(m)=2$ for $m > 1$. Thus we get $R(m,0) \le m(m-1)+s(m)$ for $m \ge 0$.
Now we do the same thing, increasing the number of lines step by step. If we add a line to an original configuration with $m$ circles and $n-1$ lines, then that new line has at most $2m$ common points with all the $m$ circles and at most $n-1$ common points with the other $n-1$ lines.
Thus we get $R(m,n) \le R(m,n-1) + (2m + n-1) + 1$.
Again, by induction over $n$, this leads to $R(m,n) \le R(m,0) + 2mn + \frac{n(n+1)}{2} = m(m-1) + s(m) + 2mn + \frac{n(n+1)}{2}$.
Now I claim that the inequality is really an equality.
Obviously, any configuration of $m$ circles and $n$ lines can be obtained by starting with the empty plane and first adding circles, one by one and then adding lines, one by one. In each step, if the used bound on $R(m,0)$ and $R(m,n)$, resp., is sharp, depends on exactly 2 things:
A) That the number of common points between the new element and the old elements is really the maximum value (1 for a line intersectiong another line, 2 for a circle intersecting another circle or line), and
B) that those common point are all different.
So in order prove the equality, we need to construct $m$ circles and $n$ lines that fullfill A) and B).
Select a point M in the plane. Choose $m$ different circles with radius equal to 1 in the plane, whose midpoints have distance at most $\frac13$ to M. These circles
1) all intersect in 2 points, and
2) all contain M as an inner point.
1) makes A) true for the circles. It should also be easy to see B) can be fullfilled, as the set from which midpoints can be chosen is a 2-dimensional object (a filled circle), while the condition that a new circle should not go through an existing intersection point excludes only a 1-dimensional set of points.
2) makes sure that :
3) we have a small circle $C$ with some radius $\epsilon >0$ around M that is inside all the circles.
Take any set of $n$ lines in general position (no 2 parallel, no 3 intersecting in 1 point). Apply a centric compression to the lines such that all the intersections are inside a circle of radius $\epsilon$. Now translate the compressed lines such that all the intersections are inside the circle $C$.
This means all the intersections between the lines are different from all the intersections between the circles. In addition, each line contains a point (any of it inisde $C$) that is inside each circle (see 3) above), so each line will intersect with each circle in exactly 2 points.
What needs to be proven to verify condition B) above is that those line vs. circle intersections are all different from the existing circle vs. circle intersections. They might not actually be, but we have one more degree of freedom to exploit: If we rotate the configuration of lines around M, all their intersections stay inside $C$. But only a finite number of rotation angles create invalid configurations where a line vs. circle intersection is equal to an existing circle vs. circle intersections, so one rotation angle can be found where this is not the case and condition B) is fullfilled.
This means we can find a configuration where all the lines and circles intersect in the maximum number of ways in all different points, so in the estimations for $R(m,n)$ we would always get equality in each step.
This concludes the proof that
$$R(m,n) = m(m-1) + s(m) + 2mn + \frac{n(n+1)}{2}$$
where $s(m)=1$ for $m=0$ and $s(m)=2$ for $m \ge 1$.