Signature of the manifold of the multiple fibrations over spheres We can define the signature of a manifold in $4k$ dimensions.
1) If I understand correctly, the signature  $\sigma$ of the manifold of the product space of spheres would always be zero:

$$\sigma(S^n \times S^m \times S^p \times S^q \times \dots )=0$$

Yes?
2) Am I correct that the signature of the manifold of the (multiple)
 fibrations over spheres will always be zero?
$$S^n \hookrightarrow X_1 \rightarrow  S^m$$
$$X_1 \hookrightarrow X_2 \rightarrow S^p$$
$$X_2 \hookrightarrow X_3  \rightarrow S^o$$

$$\sigma(X_j)=0?$$ 
  (for $j=1,2,3,...$, $m,n,o$ are positive integers.) 

If I am wrong, is this statement true in 4-dimensions?
Or, can one give one or two simplest counterexamples in 4-dimensions? (or other dimensions?)
See also this post.
 A: The total space of a fiber bundle over a sphere that isn't a circle has zero signature by a  result of 
Chern, Hirzebruch, and Serre that the signature is multiplicative when the fundamental group of the base acts trivially on the cohomology of the fiber.
A: The signature of an orientable bundle over a sphere with closed fiber vanishes. 
Here is an argument via Novikov additivity. Any bundle over $S^k$ with fiber $N^{4n-k}$ is obtained by gluing two copies of $N \times D^k$ (clutching construction) along $N \times \partial D^k$. Now the signature of $N \times D^k$ is zero, because the homology of the boundary, $N \times \partial D^k$, maps surjectively to the homology of $N \times D^k$.
Novikov additivity says that the signature of the original bundle is the sum of the signatures of the two pieces; hence the signature vanishes (without the need for the general Chern-Hirzebruch-Serre result cited above). 
Note that this argument fails if $N$ is not closed, because Novikov additivity requires that you glue along closed components of the boundary. A simple example of this failure would be an oriented $D^2$ bundle over $S^2$. The signature is the sign of the Euler class of the bundle.
