Let $M$ be a smooth manifold, $E\subseteq TM$ be an integrable distribution and $\pi:TM\to Q=\frac{TM}{E}$ the quotient bundle with $\dim Q=q$. Bott showed that \begin{equation} Pont^k(Q)=0 \qquad \forall k>2q \label{dd} \end{equation} and togehter with the proposition that $Pont(Q)=Pont(Q')$ for homotopy equivalent distributions $E$ and $E'$ it follows that the Pontryagin classes in the above equation already vanish when $E$ is homotopy equivalent to an integrable distribution.
I don't understand one step in the argument.
One shows that there is exists a partial connection $\nabla:\Gamma(E)\times \Gamma(Q)\to\Gamma(Q)$ defined by \begin{equation} \nabla_X(Z)=\pi[X,\tilde Z] \end{equation} where $Z=\pi(\tilde Z)$. One can show that the corresponding 'partial ' curvature form $k$ vanishes \begin{equation} k(X,X')(Z)=\nabla_X\nabla_{X'}Z-\nabla_{X'}\nabla_X Z -\nabla_{[X,X']}Z=0 \end{equation}
How does this show that the Pontryagin classes vanish beyond the order $2q$?