composition of stereographic projections is inversion through the ball - a geometric way $S^{n}$ be the unit sphere in $\mathbb{R}^{n+1}$.  Let $\pi_N$ be stereographic projection from the sphere without the north pole on to $\mathbb{R}^{n}$ and let $\pi_S$ be defined similarly using the south pole.
On $\pi_N(\pi_S^{-1}(\mathbb{R^n}))=\mathbb{R^n-0}$, one can take $\pi_S \circ \pi_N^{-1}$.  By messy algebra, I showed that this is the inversion through the unit sphere $S^{n-1} \subset \mathbb R^{n}$ sending $x \in \mathbb{R}^{n} \mapsto x/|x|^2$. 
In full detail, I showed that that under this correspondence $z \mapsto z \cdot 2/(1+|z|^2)+(|z|^2-1)/(|z|^2+1)(0,....,0,1) \mapsto z  \cdot (2/(1+|z|^2))/(2 |z|^2 /(1+|z|^2))=z/|z|^2$
Is there a geometric way of seeing this using just inner products and such. i.e. I don't want to have to use the quadratic forumula like I did (I am trying to do this along the lines of the symmetry lemma given in Axler's Harmonic function theory. You don't have to though.)
 A: The points, its projections and the poles all reside in some 2D cross section of space, so it suffices to do the problem in 2D. I assume you're working with the unit circle.
Let's draw a picture and label some points. Say $N$ is the north pole, $S$ is the south pole, $O$ is the origin, $Q$ is a point on the unit circle (without loss of generality in quadrant IV), with stereographic projections given by $\pi_S(Q)=P$ and $\pi_N(Q)=R$.
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We will show the triangles $\triangle NOR$ and $\triangle POS$ are similar by showing they have the same angle measures (in order). Similarity implies an equality of ratios $[PO:OS]=[NO:OR]$. Since both of the line segments $NO$ and $OS$ are length $1$, this proves $PO$ and $OR$ are reciprocal lengths.
By the "angle inscribed in semicircle" theorem, $\angle NQS$ is a right angle. Clearly $\angle NOR$ is also a right angle. Note that the angles $\angle NRO$ and $\angle PRQ$ are equal, and complementary to $\angle ONR$ and $\angle OPS$ which are in turn complementary to $\angle NSP$.
