How to prove that automorphisms of $\mathbb P^n$ arise from linear maps in $\mathbb A^{n+1}$?

Let $\varphi\in \mathrm{Aut}(\mathbf{P}^n_k)$ be an automorphism.

Then $\varphi$ induces an automorphism on the level of the Picard group of $\mathbf{P}^n_k$. Now, you might know that $\mathrm{Pic}(\mathbf{P}^n_k)\simeq \mathbb {Z}$ via the degree map on the level of invertible sheaves: the isomorphism assigns to each invertible sheaf $\mathscr{L}$ its degree $\deg \mathscr{L}$.

It follows that the automorphism $\varphi^* :\mathrm{Pic}(\mathbf{P}^n_k)\longrightarrow \mathrm{Pic}(\mathbf{P}^n_k)$ must send a generator onto a generator, hence $\varphi^*(1)$ is either $1$ or $-1$. In sheaf-theoretic terms, one has that $\varphi^*(\mathscr O (1))$ is either $\mathscr{O}(1)$ or $\mathscr{O}(-1)$.

However, one has that

$$\dim H^0(\mathbf{P}^n,\mathscr O(1))= n +1 \neq 0 =\dim H^0 (\mathbf{P}^n, \mathscr{O}(-1))$$

which contradicts the fact that $\varphi$ is an isomorphism, hence inducing $H^i(\mathbf P^n,\mathscr{F})\simeq H^i (\mathbf{P}^n,\varphi^*\mathscr F)$

So we have $\varphi^* \mathscr (1) = \mathscr O (1)$; this implies that $\varphi^*$ induces a linear automorphism on $H^0(\mathbf{P}^n_k,\mathscr{O}(1)$, which amounts to a size $(n+1)$ square matrix $\Phi: k^{n+1}\longrightarrow k^{n+1}$.

What does this mean in terms of $\varphi$?. Let us introduce homogeneous coordinates on $\mathbf {P}^n$: these are just $n+1$ global sections $\sigma_0,\ldots ,\sigma_n\in H^0 (\mathbf{P}^n_k,\mathscr{O}(1))$ with no common zeroes and one represents each $p\in \mathbf {P}^n$ with

$$[\sigma_0(p):\cdots :\sigma_n(p)]\in \mathbf{P}(H^0(\mathbf{P}^n,\mathscr{O}(1)))$$

So, in coordinates, we have that

$$\varphi(p) = [\varphi^* \sigma_0 (p) :\cdots : \varphi^* \sigma_n (p)] = [\Phi (\sigma_0(p),\ldots ,\sigma_n(p)] $$

The function field of $\Bbb{P}^n$ is the same as $\Bbb{A}^n$ : $k(x),x=(x_1,\ldots,x_n)$.

The difference is that $\Bbb{P}^n$ is complete : if $f\in k(x)$ then its divisor of zeros poles on $\Bbb{P}^n$ is a weigthed sum $\sum_j n_j Z(p_j)$ of irreducible hypersurfaces $Z(p_j)$ each being the vanishing set of an (irreducible) homogeneous polynomial $p_j$ of degree $\deg(p_j)$ and $\sum_j n_j \deg(p_j) = 0$.

In other words : if $p_1,\ldots,p_j$ are homogeneous polynomials and $n_j$ are integers and $\sum_j n_j\deg(p_j)=0$ then $\prod_j p_j^{n_j}$ is in the function field of $\Bbb{P}^n$ and every element of the function field appears this way.

The group of principal divisors is $Prin=\{\sum_j n_j Z(p_j),\sum n_j \deg(p_j)=0\}$, the whole group of divisors is $Div= \{ \sum_j n_j Z(p_j)\}$, the quotient $Div/Prin$ is just given by $\sum_j n_j Z(p_j)\mapsto \sum_j n_j \deg(p_j)$ ie. $Div/Prin \cong \Bbb{Z}$.

If $\varphi \in Aut(k(x))$ then $\varphi \in Aut(Div),\in Aut(Prin),\in Aut(Div/Prin)= Aut(\Bbb{Z})$ and it is not hard see $\varphi$ cannot be $\Bbb{Z}\mapsto -\Bbb{Z}$ so it is the identity on $\Bbb{Z}$, ie. $\varphi$ sends $Z(p_j)$ to $\sum_i n_i Z(q_i)$ with $\deg(p_i)=\sum_i n_i \deg(q_i) $.

But if $\varphi \in Aut(\Bbb{P}^1)$ then it is more than that : $\varphi$ must send the zeros of $f$ to the zeros of $f\circ \varphi$, which means it must send $Z(p_j)$ to $\sum_j n_jZ(q_j)$ with $n_j \ge 0$ and by the preceding if $\deg(p_j)=1$ it must send $Z(p_j)$ to $Z(q_j)$ with $\deg(q_j)=1$.

Taking the basis of degree 1 homogeneous polynomials which are the usual chart for $\Bbb{P}^n$ we are done : $\varphi\in GL_{n+1}(k)$ and $Aut(\Bbb{P}^n)=PGL_n(k)$.