You can induct on $n$. Let $f:X\to\mathbb{P}^n$ be finite and étale.
If $H$ is a hyperplane in $\mathbb{P}^n$, there is a trivialization $\phi:f^{-1}(H)\simeq H\times F$($F$ , for a finite ) $F$, by the induction hypothesis.
If $L$ is any line in $\mathbb{P}^n$, $f^{-1}(L)$ is a finite disjoint union of $\mathbb{P}^1$'s, and you can label the components by elements of $F$ using the trivialization at the any point of $L\cap H$ (in case $L\subset H$, otherwise there is only one).
Now any fiber $f^{-1}(x)$ f^{-1}(x)$,$x\in \mathbb{P}^n$, is identified with$F$through the labeling of the components of$f^{-1}(L)$, for any line$L$through$x$(this doesn't depend on the line , through$x$, their space being connected). 2 to index --> to label ; tentative grammar corrections You can induct on$n$. Let$f:X\to\mathbb{P}^n$be finite and étale. If$H$is a hyperplane in$\mathbb{P}^n$, there is a trivialization$\phi:f^{-1}(H)\simeq H\times F$($F$finite) by the induction hypothesis. If$L$is any line in$\mathbb{P}^n$,$f^{-1}(L)$is a finite disjoint union of$\mathbb{P}^1$'s, that and you can index label the components by elements of$F$using the trivialization at the point$L\cap H$. Now any fiber$f^{-1}(x)$is identified with$F$through the indexing labeling of the components of$f^{-1}(L)$,$L$for any line$L$through$x$(this doesn't depend on the line, their space being connected). 1 You can induct on$n$. Let$f:X\to\mathbb{P}^n$be finite and étale. If$H$is a hyperplane in$\mathbb{P}^n$, there is a trivialization$\phi:f^{-1}(H)\simeq H\times F$by induction hypothesis. If$L$is any line in$\mathbb{P}^n$,$f^{-1}(L)$is a finite disjoint union of$\mathbb{P}^1$'s, that you can index by$F$using the trivialization at the point$L\cap H$. Now any fiber$f^{-1}(x)$is identified with$F$through the indexing of the components of$f^{-1}(L)$,$L$any line through$x\$ (this doesn't depend on the line, their space being connected).