Consider the canonical bundle $ K_{ \mathbb{P}^{n} } = \bigwedge^n T_{\mathbb{P}^{n}}^* $, which is the line bundle whose sections are holomorphic forms of top degree, expressible locally as $ f(z_1 , \dots , z_n ) dz_1 \wedge \dots \wedge z_n $. For $ \omega $ an $n$-form on $ \mathbb{P}^n $, write $ ( \omega ) $ for the corresponding divisor.
Questions :
Why does the canonical class $ K_{ \mathbb{P}^{n} } $ of $ \mathbb{P}^n $, which is the class equivalence of $ ( \omega ) $ in $ \mathrm{Pic} ( \mathbb{P}^{n} ) = \{ \ \text{line bundles on} \ \mathbb{P}^n \ \} = \{ \ \text{divisors on} \ \mathbb{P}^n \ \} / \sim \ $ checks the following property : $$ K_{ \mathbb{P}^{n} } = -(n+1) H = \mathcal{O}_{ \mathbb{P}^{n} } (-n-1) $$ such that $ H $ the hyperplane of $ \mathbb{A}^n $ at infinity.
Thanks in advance for your help.
The quickest argument is to use the Euler sequence: $0\to \Omega^1\to O(-1)^{n+1}\to O\to 0$. Taking maximal exteriors you get $\Lambda^n\Omega^1=O(-n-1)$.