Chern classes of complex vector bundle I'm reading characteristic classes form the book Differential forms in Algebraic Topology by Bott and Tu. The Chern classes are defined as follows:
$E\xrightarrow{\rho} M$ is a vector bundle and $E_p$ be the fiber over $p$. Then $P(E)$, the projectivization of $E$ is a vector bundle with fiber $P(E_p)\colon=\{\text{1-dim subspaces of $E_p$}\}$ over $\ell_p\in P(E)$. It's then discussed that the first Chern class $x$ of the dual of the universal subbundle over $P(E)$ restricted to a fiber is the first Chern class of the fiber. Therefore the $1,x,\ldots,x^{n-1}$ are global classes on $P(E)$, whose restriction to each fiber $P(E)$ freely generates the cohomology of the fiber and hence by Leray-Hirsch $H^*(P(E))$ is a free module over $H^*(M)$ with basis $\{1,x,\ldots,x^{n-1}\}$ and then the Chern classes are the coefficients in the expression for $x^n$.
Also by Leray-Hirsch, $H^*(P(E))\cong H^*(M)\otimes H^*(Fiber)$. The fiber $P(E_p)$ has cohomology $\frac{\mathbb{R}[x]}{x^n}$, as $E_p$ is a vector space of complex dimension $n$. My question is, why $x^n\neq 0$.
Note: for the trivial bundle $M\times V$, $P(E)=M\times P(V)$ and by Künneth formula $H^*(P(E))\cong H^*(M)\otimes H^*(P(V))$ and then $x^n=0$ as $H^*(P(V))=\frac{\mathbb{R}[x]}{x^n}$. What is the difference in the general case and the trivial bundle?
 A: Bertram already mentioned this in the comments but I thought I'd write an answer for completness's sake.
The Leray-Hirsch theorem says that $H^{*}P(E)\cong H^{*}M\otimes H^{*}(Fiber)\ \ $ $\textit{as $H^{*}M$ modules}$.
So if $x$ is the first chern class of the tautological line bundle over $P(E)$, there's no reason to expect $x^{n}=0$, even if its pullback to the fiber satisfies that equation. Because if the fiber "twists around," there's no reason to expect that the behavior of a class on one fiber is indicative of its behavior globally. 
On the other hand, when the bundle is trivial, the Kunneth formula (in this setting) tells you that $H^{*}P(E)\cong H^{*}M\otimes H^{*}(Fiber)\ \ $ $\textit{as rings}$.
Here's another fun example to illustrate the difference, where you can even draw pictures! You'll have to use cohomology over $\mathbb{Z}/2$ and you'll be calculating Stiefel-Whitney classes instead of chern classes, but the procedure is exactly the same: Leray-Hirsch, read off coefficients from a basis. Form $E\rightarrow S^1$ by taking the Mobius strip as a vector bundle over $S^1$ plus a copy of the trivial line bundle. Then $P(E)$ will be a familiar non-orientable surface (Klein bottle) whose cohomology ring you can compute from a delta-decomposition. Comparing it to the cohomology ring of the torus (which would be $P(E)$ for the trivial rank 2 bundle) where every positive degree class squares to zero, you can see exactly how the "twist" in the Klein bottle allows one of those classes to no longer square to zero! 
