Why is the etale fundamental group of $\mathbb{P}^n$ trivial?

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For $n=1$, one just uses the Riemann-Hurwtiz formula. I am curious about how to do it for $n\geq 2$. I spoke to a colleague and their proof did not click with me and I've since forgotten how it goes.

I know there is an immediate way of seeing it is trivial by using the Riemann Existence Theorem, but I'd like to avoid passing to the complex side of things.

So basically, what is a direct proof that $\pi_1^{et}(\mathbb{P}^n)=1$?

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This can be proven by induction. The base case is $n=1$ which follows from the Riemann-Hurwitz formula as you describe in your post.

For the inductive step, suppose $f:X\to\Bbb P^n_k$ is a nontrivial etale cover with $X$ connected. Let $H\subset\Bbb P^n_k$ be a hyperplane, and consider $X\times_{\Bbb P^n} H\to H$, a finite etale cover of $H\cong\Bbb P^{n-1}_k$. I claim that $X\times_{\Bbb P^n} H$ is connected: $f^*\mathcal{O}_{\Bbb P^n_k}(1)$ is ample on $X$ because $\mathcal{O}_{\Bbb P^n_k}(1)$ is ample and $f$ is finite surjective (ref 0B5V), and then $X\times_{\Bbb P^n_k} H$ is the support of this ample divisor inside a nice variety of dimension $\geq 2$ hence connected (ref MSE or Hartshorne corollary III.7.9). By the inductive hypothesis, $X\times_{\Bbb P^n} H\to H$ is of degree 1, hence $X\to\Bbb P^n$ must also be of degree one, or an isomorphism.