Now let $\pi:\Bbb{C}^{n+1}\backslash\{0\}\to\Bbb{CP}^n$ be the natural projection, and $\mathcal{O}(-1)$ be its tautological line bundle, defined by $\{([x],l)\in\Bbb{CP}^n\times\Bbb{C}^{n+1}\mid l\in x\}$. Then consider its pullback by $\pi$, that is $$ \pi^*(\mathcal{O}(-1))\subset\Bbb{C}^{n+1}\backslash\{0\}\times\mathcal{O}(-1) $$
I need to show it's a trivial line bundle, is it really right?
If you write the definition of pullback bundle
$$ \pi^*(\mathcal{O}(-1)) = \{ (x,v)\in \mathbb{C}^n\setminus\{0\} \times \mathcal{O}(-1)| \pi(x) = \pi_L(v)\} $$ where $\pi_L:\mathcal{O}(-1)\rightarrow \mathbb{CP}^n$ is the bundle projection, you realize that $\pi(x)=[x]=\pi_L(v)$ implies that $v$ is an element of $\mathbb{C}^{n+1}$ in the line spanned by $x$. This means that the fiber of the pullback bundle $\pi^*(\mathcal{O}(-1))$ over $x$ is precisely the set $\{(x,(x,\lambda x)):\lambda\in\mathbb{C}\}$. This happens globally and there is an obvious vector bundle isomorphism with $\mathbb{C}^{n+1}\setminus\{0\} \times \mathbb{C}$.