Easy proof of topological property of Zoll manifolds It is known that the cohomology ring of a Zoll manifold---a riemannian manifold all of whose geodesics are periodic with the same minimal period---must be the same as the cohomology ring of a compact rank one symmetric space (see Besse's book Manifolds all of whose geodesics are closed for references). 
Is there a simple and elementary proof of the following much weaker property?
The first Betti number of a Zoll manifold is equal to zero. 
Addendum.
The comment by Thomas Richard got me thinking and here is something that should lead to a proof that the fundamental group of a Zoll manifold is either trivial or isomorphic to $\mathbb{Z}_2$:
Any two prime closed geodesics in a Zoll manifold are homotopic. Indeed, if $v_x$ is a unit vector tangent to a geodesic $\gamma$ and $w_y$ is a unit vectors tangent to a geodesic $\sigma$, then a continuous path on the unit tangent bundle joining these two unit vectors, taken as the initial conditions of prime closed geodesics, will define a homotopy between  $\gamma$ and $\sigma$.
Note that there is at least one closed geodesic representing each non-trivial homotopy class of loops, but the geodesic doesn't have to be prime. Still ...  
 A: As asked by @alvarezpaiva, I repost my remark as an answer.
After his addendum and answer showing that the fundamental group (if non-trivial) has one generator, given by a prime closed geodesic, you just have to observe that this geodesic is homotopic to itself with reversed orientation.
Hence the fundamental group has order at most $2$.
A: I think I have it: The fundamental group of a Zoll manifold is a finite cyclic group.
Proof. In a compact riemannian manifold every non-trivial element of the fundamental group is represented by a closed geodesic. As was remarked in the addendum to the OP, all prime closed geodesic in a Zoll manifold are homotopic and, since every closed geodesic is an iterate of a prime geodesic, this implies that the fundamental group---if non-trivial---has just one generator. 
In order to show that it is a finite group, consider the universal cover of the Zoll manifold, which is itself a Zoll manifold and hence compact. OK this part is still a bit fuzzy in my mind ... Is there a quick argument for proving that the universal cover of a Zoll manifold is compact? This was probably the crux of the matter from the beginning anyway.
