$$B_{n-t} \rightarrow B_{n-t-1} \rightarrow \cdots \rightarrow B_1$$ is a trivial Bott tower. Therefore $B_{\ell} = ({\mathbb{C}P}^1)^{\ell}$ for $\ell=1, \ldots, n-t$.

Let $s$ be the cohomological complexity of $M$. Then it is clear that $t\geq s$ in general. Suppose $t>s$. Since the twist number of $M$ is $t$, we have $$H^\ast(B_n) = {\mathbb{Z}}[x_1, \ldots, x_n] / <x_j(x_j - f_j) \mid j=1,\ldots,n>,$$ where $$f_j = \left\{\begin{array}{ll}
    0 & \text{for $1 \leq j \leq n-t$} \\
    \sum_{i=1}^{j-1} \c{j}{i}x_i & \text{for $ n-t < j \leq n$}. \\
\end{array}\right.$$

Since the cohomological complexity of $B_n$ is $s$, there is an isomorphism $$\psi : H^\ast(B_n) \rightarrow {\mathbb{Z}}[y_1, \ldots, y_n] / <y_j(y_j - g_j) \mid j=1,\ldots,n>,$$ where $$g_j = \left\{\begin{array}{ll}
    0 & \text{for $1 \leq j \leq n-s$}, \\
    \sum_{i=1}^{j-1} \d{j}{i}y_i & \text{for $n-s< j \leq n$.}
\end{array}\right.$$

**We claim that there exists $m$ ($n-t < m \leq n$) such that $f_m
\equiv 0 \mod 2$ and $f_m^2 = 0 \in H^\ast(B_{m-1})$.**If the claim is true, then we can write as $f_m + 2w = 0$ for some $w \in H^2(B_{m-1})$. Therefore, $$\begin{aligned}
c(\gamma^w \oplus \gamma^{f_m + w}) &= (1+w)(1+f_m+w) = 1+ (f_m + 2w) - \frac{f_m^2}{4} \\
& = 1\end{aligned}$$ Thus, by Proposition \[proposition:trivial bundle\], 