Canonical sheaf of $\mathbb{P}^n$ by 8.20.1 in Hartshorne

Question

Hartshorne use the dual Euler exacte sequence for $X=\mathbb{P}^n_k$:
$$ 0\to \Omega_X\to \mathcal{O}_X(-1)^{\oplus n+1}\to \mathcal{O}_X\to 0 $$
then taking $\text{det}$ (highest exterior powers) we get
$$ \text{det}(\Omega_X)\otimes\text{det}(\mathcal{O}_X)\simeq\text{det}(\mathcal{O}_X(-1)^{\oplus n+1}) $$
Then we have $\text{det}(\Omega_X)=\omega_X$ the canonical sheaf, $\text{det}(\mathcal{O}_X)=\wedge^1\mathcal{O}_X=\mathcal{O}_X$ so the LHS is $\omega_X$. For the RHS it is less clear to me. I guess we have
$$ \text{det}(\mathcal{O}_X(-1)^{n+1})=\wedge^{n+1}(\mathcal{O}_X(-1)^{n+1}) $$
but then? How to get $\mathcal{O}_X(-n-1)$?

Answer 1

Ok thanks to Sasha all is clear: $$ \det(\mathcal{O}_X(-1)^{\otimes n+1})=\det(\mathcal{O}_X(-1)\otimes \ldots\otimes\mathcal{O}_X(-1))=\det(\mathcal{O}_X(-1))\oplus\ldots\oplus\det(\mathcal{O}_X(-1)) $$ But $\det(\mathcal{O}_X(-1))=\wedge^1\mathcal{O}_X(-1)\simeq\mathcal{O}_X(-1)$ so $$ \det(\mathcal{O}_X(-1)^{\otimes n+1})=\mathcal{O}_X(-1)\otimes\ldots\otimes\mathcal{O}_X(-1)=\mathcal{O}_X(-n-1) $$