Prove that $n$ circles in general position (no three circles share the same point, and every two circles intersect each other), divide a plane in $n^2 -n +2$ areas.
Proof for circles in a plane: $P(n)=n^2-n+2 \implies P(1)=2$
Since one circles divides a plane in two parts, the statemet is true for$\;n_0=1.$
By the induction assumption: $P(k)=k^2-k+2.$
We have to prove: $\;P(k+1)=(k+1)^2-(k+1)+2=k^2+k+2$
If we add the $(k+1)th$ circle to the set of $k$ circles, then those $k$ circles intersect the circle (added to the set) in $2k$ points.
Those $2k$ points form $2k$ consecutive arcs, and every new arc divides the former part (arc) of the circle, in two parts. Therefore, by adding the $(k+1)th$ circle to the set of $k$ intersecting circles, we got $2k$ more areas. $$\implies P(k+1)=P(k)+2k\;\implies P(k+1)=(k^2-k+2)+2k=k^2+k+2.$$ $$\; Q.E.D$$
But what happens in a $3$-$D$ space when the intersection of two spheres is a circle?