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0\to TM \to T\mathbb{P}^n \to \mathcal{O}(n+1)|_M \to 0
0\to TM \to T\mathbb{P}^n \to \mathcal{O}(n+1)|_M \to 0
</math>
</math>
we get that <math>c(T\mathbb{P}^n) = c(TM)(1+(n+1)\omega)</math>. But this implies that
<math>
c(TM) = 1+c_1(TM)+\ldots = \frac{(1+\omega)^{n+1}}{1+(n+1)\omega}
</math>
from which one easily concludes that <math>c_1(TM) = c_1(K_M) = 0</math>.