pullback of twisting sheaf

Question

Let $[k]: \mathbf{P}^n \to \mathbf{P}^n, [x_0:\ldots:x_n] \mapsto [x_0^k:\ldots:x_n^k]$ be a morphism. (Why) do we have $[k]^*\mathcal{O}_{\mathbf{P}^n}(1) \cong \mathcal{O}_{\mathbf{P}^n}(k)$?

Answer 1

This is easy to see using the cocycle description of $\mathcal{O}(k)$, namely they are given by $\eta_{ij} = \dfrac{t_j^k}{t_i^k}$, which is $[k]^\#$ applied to the cocycle for $\mathcal{O}(1)$.

Answer 2

This is easy if you read Hartshorne GTM 52, Theorem II.7.1.

Because $[k]$ is defined by $x_0^k,\ldots,x_n^k$, which are the global sections of $\mathcal{O}_{\mathbf{P}^n}(k)$.