I've read that $\text{Br} \mathbb{P}^n_k$ (here $\text{Br}$ is the cohomological Brauer group, i.e. $H^2_{ét}(,\mathbb{G}_m)$) is just isomorphic to $\text{Br} k$. As proof of this fact seems to be not so easy in the general case, but there should be a simple and conceptual proof when $k$ has characteristic zero. Does anyone know this simple proof?

I don't think the assumption of characteristic zero simplifies the proof a great deal. However, it does allow us to avoid having to give a more involved proof for the $p$power torsion (where $p = char (k)$) so I will assume that below. Firstly, by Proposition 1.4 of Grothendieck's "Groupe de Brauer II", $H^2(X, \mathbb{G}_m)$ is torsion for a smooth variety $X$ so we may use the Kummer sequence $$ 1 \to \mu_r \to \mathbb{G}_m \stackrel{r}{\to} \mathbb{G}_m \to 1 $$ of etale sheaves on $X$ to compute the $r$torsion for all $r$ and hence compute all of $Br(X)$. The long exact sequence of etale cohomology on $X$ gives an exact sequence $$ H^1(X,\mathbb{G}_m) = Pic(X) \stackrel{d}{\to} H^2(X, \mu_r) \to Br(X) \stackrel{r}{\to} Br(X) $$ so we need to compute the cokernel of $d$. Since $H^0(\mathbb{P}_{\bar{k}}^n,\mu_r) = \mu_r$, $H^1(\mathbb{P}_{\bar{k}}^n, \mu_r) = 0$, and $H^2(\mathbb{P}_{\bar{k}}^n, \mu_r)) = \mathbb{Z}/r$, the HochschildSerre spectral sequence gives an exact sequence $$ 0 \to H^2(Gal(\bar{k}/k),\mu_r) \to H^2(\mathbb{P}_k^n, \mu_r) \to H^0(Gal(\bar{k}/k), \mathbb{Z}/r) \to 0 .$$ The map $\mathbb{Z} = Pic(\mathbb{P}^n) \to H^2(\mathbb{P}^n, \mu_r) \to H^2(\mathbb{P}_{\bar{k}}^n, \mu_r) = \mathbb{Z}/r$ is surjective so it follows that $H^2(Gal(\bar{k}/k),\mu_r) = Br(k)[r]$ maps isomorphically onto $Cokernel(d) = Br(\mathbb{P}^n)[r]$. Since this is true for any integer $r$ it follows that the map $Br(k) \to Br(\mathbb{P}^n)$ is an isomorphism. 


Let $X$ be an $n$dimensional projective space over a field $k$. Let $k_s$ be a separable closure of $k$, and $X_s$ the base change of $X$ to $k_s$. The algebraic part $\textrm{Br}_1(X)$ of the Brauer group of $X$ is defined as $\textrm{Br}_1(X) = \ker(\textrm{Br}(X) \rightarrow \textrm{Br}(X_s))$ and sits inside a short exact sequence $$0 \rightarrow \textrm{Br}(k) \rightarrow \textrm{Br}_1(X) \rightarrow H^1(G_k,\textrm{Pic}(X_s)) \rightarrow 0$$ given by the HochschildSerre spectral sequence. Since $\textrm{Pic}(X_s) = \mathbf{Z}$, it follows that we have a canonical isomorphism between $\textrm{Br}_1(X)$ and $\textrm{Br}(k)$. So now we are reduced to showing that $\textrm{Br}(X_s)$ is trivial. I do not how to do this at the moment. 


Consider the commutative diagram \begin{array}{ccc} \operatorname{Br}k & \xrightarrow{f_1} & \operatorname{Br} \mathbb{P}_{k}^{n} \\ \scriptsize{f_2}\ \downarrow & \swarrow \scriptsize{f_3}& \downarrow \scriptsize{f_4}\\ \operatorname{Br}\mathbb{A}_{k}^{n} & \xrightarrow[f_5]{} & \operatorname{Br}k(x_{1},\dotsc,x_{n}) \end{array} induced by pullback. By [Milne, Etale Cohomology, Corollary IV.2.6], $f_{4}$ is injective; hence $f_{3}$ is injective. If $k$ has characteristic zero, $f_{2}$ is an isomorphism by [AuslanderGoldman, "The Brauer group of a commutative ring", Proposition 7.7]; thus $f_{3}$ is surjective. EDIT: See also Proposition 1.5 of Reineke and Schroer's paper "Brauer groups for quiver moduli", http://arxiv.org/abs/1410.0466. 

