I'm working on this problem from section 4.2 of Hatcher's Algebraic Topology:
Show that if $S^k\to S^m\to S^n$ is a fiber bundle then $k=n-1$ and $m=2n-1$.
By counting dimensions, we have $m=n+k$, and if I assume that $k<n$ the problem is not too difficult. But, this assumption on $k$ is not stated in the problem. Is there a clean way to show that $k<n$?
Suppose $k > 0$.
The long exact sequence in homotopy gives
$$\dots \to \pi_n(S^{n+k}) \to \pi_n(S^n) \to \pi_{n-1}(S^k) \to \pi_{n-1}(S^{n+k}) \to \dots$$
As $n < n + k$, $\pi_n(S^{n+k}) = 0$ and $\pi_{n-1}(S^{n+k}) = 0$, so $\pi_n(S^n) \cong \pi_{n-1}(S^k)$. We know $\pi_n(S^n) \cong \mathbb{Z}$, so $\pi_{n-1}(S^k) \cong \mathbb{Z}$ and hence $n - 1 \geq k$, i.e. $n > k$.
For $k = 0$, the only fiber bundle (or even fibration) is $S^0 \to S^1 \to S^1$ - see this question. In this case we also have $k = n -1$ and $m = 2n-1$.