Can I build fibre bundles over bundles ad infinitum? Given a manifold, I can construct a fiber bundle over it, e.g. the tangent bundle. This fiber bundle is also a manifold, so I can construct a bundle over it, e.g. another tangent bundle. This is again a manifold... and so on and so forth.
I was wondering: Is there an end to this construction? Does it stop to be "interesting" from the mathematical point of view at some point?
 A: Of course, you can iterate the tangent bundle construction infinitely and obtain "new" bundles in this way. The question of when it stops being "interesting" from the mathematical viewpoint is a different matter, however. For example: any vector bundle is homotopy equivalent to the base space. As such, if you would ask an algebraic topologist, they might tell you that there is no meaningful distinction between $M$ and $T^kM$, as topological spaces. For example, there is a bijection between the isomorphism classes of vector bundles over $M$, and isomorphism classes of vector bundles over $T^kM$, where $T^kM$ is the $k$-iterate of the tangent bundle. The cohomology and homotopy groups of these spaces are also isomorphic. (The same holds for any "tower" of vector bundles over a base space $M$).
Indeed, the general significance of $TM$ is not that it offers a new topological space to study, but that it is a vector bundle whose sections are vector fields, which allow us to do calculus on the manifold $M$. So I, for one, would argue that it very rarely happens that higher iterates of this construction are "mathematically interesting", although I'm by no means an expert and if someone disagrees then I'd be happy to hear it.
One exception to this occurs in classical mechanics. Here, one considers $M$ the state space of a system, and $T^*M$ the phase space. Then, the object $T(T^*M)$ is in fact important.
As a final note: when one replaces vector bundles by fibre bundles more generally, then yes, one can also obtain topologically interesting spaces by just iterating this procedure.
