One may deduce the formula $n^{2}-n+2$ as follows: Start with $m$ circles already drawn on the plane with no two of them tangent, none of them being inside another and no three of them concurrent. Then draw the $m+1$ circle $C$ so that is does not violate the propeties stated before and see how it helps increase the number of regions. Indeed, we can see that that $C$ intersects each of the remaining $m$ circles at two points. Therefore, $C$ is divided into $2m$ arcs, each of which divides in two a region formed previously by the first $m$ circles. But a circle divides the plane into two regions, and so we can count step by step ($m=1,2,\cdots, n$) the total number of regions obatined after drawing the $n$-th circle. That is,
$$
2+2(2-1)+2(3-1)+2(4-1)+\cdots+2(n-1)=n^{2}-n+2
$$
Since $n^{2}-n+2\ge (n^{2}+n+2)/2$ for $n\ge 1$ the answer is affirmative.
ADDENDUM: An easy way to see that the answer to my question is affirmative without finding a formula may be as follows: Suppose that $l_{n}$ is the maximum number of regions into which the plane $\mathbb{R}^{2}$ can be divided by $n$ lines, and that $c_{n}$ is the maximum number of regions into which the plane can be divided by $n$ circles.
Now, in the one-point compactification $\mathbb{R}^{2}\cup\{\infty\}$ of the plane, denoted by $S$ (a sphere), the $n$ lines become circles intersecting all at the point $\infty$. Therefore, these circles divide $S$ into at least $l_{n}$ regions. Now, if we pick a point $p$ in the complement in $S$ of the circles and take the stereographic projection through $p$ mapping onto the plane tangent to $S$ at the antipode of $p$ we obtain a plane which is divided by $n$ circles into at least $l_{n}$ regions. Therefore, $l_{n}\le c_{n}$.
Moreover, from this we can see that the plane and the sphere have equal maximum number of regions into which they can be divided by circles.