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?