Shortest path on a sphere

I'm quite a newbie in differential geometry. Calculus is not my cup of tea ; but I find geometrical proofs really beautiful. So I'm looking for a simple - by simple I mean with almost no calculus - proof that the shortest path between two points on a sphere is the arc of the great circle on which they lie. Any hint ?

Edit: Or at least a reference ?

Here's a geometric observation that can hardly be called a "proof", but may be appealing nonetheless.

If $p$ and $q$ are distinct points of the sphere $S^{n}$, if $C:[0, 1] \to S^{n}$ is a "shortest path" joining $p$ to $q$, and if $F:S^{n} \to S^{n}$ is a distance-preserving map fixing $p$ and $q$, then $F \circ C$ is also a shortest path (because the length of $F \circ C$ is equal to the length of $C$).

Assume $q \neq -p$. If you believe there exists a unique shortest path from $p$ to $q$, it's not difficult to see that the "short" great circle arc is the only candidate: Every point not on the great circle through $p$ and $q$ is moved by some isometry of the sphere that fixes $p$ and $q$.

If you're thinking specifically of $S^{2}$, reflection $F$ in the plane containing $p$, $q$, and the center of the sphere is an isometry, and $f(x) = x$ if and only if $x$ lies on the great circle through $p$ and $q$.

(A similar argument "justifies" that the shortest path between distinct points of the Euclidean plane is the line segment joining them.)

You can show that great circle arcs are geodesics by parameterizing such an arc so that it has unit speed, and then showing that the acceleration along the arc is perpendicular to the sphere surface. (This assumes you accept that geodesics have the characteristic that their acceleration is perpendicular to the tangent plane of the surface at each point of the geodesic.)

Then uniqueness of the geodesic from a point in a direction shows it must be the great circle arc.

I realize this might not be what you seek because it doesn't connect directly to shortest paths...