Brauer group of $k$-rational scheme let $X$ be a smooth, projective and geometrically integral $k$-scheme. the Brauer group of $X$ is defined by $Br(X)=H^2_{ét}(X, \mathbb{G}_m)$. 
I'm searching for a proof of this Theorem: assume that $X$ as above and $X$ is $k$-rational, ie birational equivalent to some $\mathbb{P}^n_k$. then $Br(X)=Br(k)$.
obviously the problem can be splitted in two statements: 
1) $X,Y$ smooth, projective and geometrically integral $k$-schemes which are birationally equivalent to each other. then $Br(X)=Br(Y)$.
2) $Br(\mathbb{P}^n_k)=Br(k)$
could anybody sketch these proofs or give a reference? additionally: do we need for 1) and 2) really every of the smooth, projective and geometrically integral conditions or can it be weakened?
 A: Let me spell out how the computation for (2). We first note that we have the Kummer short exact sequence, which is 
$$ 0\rightarrow \mu_l \rightarrow \mathbb{G}_m\rightarrow \mathbb{G}_m\rightarrow 0.$$
The induced long exact sequence on etale cohomology is then 
$$ H^1(\mathbb{P}_k^n,\mathbb{G}_m) \rightarrow H^2(\mathbb{P}_k^n, \mu_l) \rightarrow H^2(\mathbb{P}_k^n,\mathbb{G}_m)\rightarrow H^2(\mathbb{P}_k^n,\mathbb{G}_m).$$
Note that $H^1(\mathbb{P}_k^n,\mathbb{G}_m)=\text{Pic}(\mathbb{P}^n_k)=\mathbb{Z}$. Then we want to compute $H^2(\mathbb{P}_k^n,\mu_l)$. For this we use the Hochschild-Serre spectral sequence. This then gives us
$H^i(G_k, H^j(\mathbb{P}^n_{\bar{k}} ,\mu_l))$ $\Rightarrow$ $H^{i+j}(\mathbb{P}_{\bar{k}}^n, \mu_l)$.
Hence we want to compute $H^0(\mathbb{P}_{\bar{k}}^n,\mu_l)$, $H^1(\mathbb{P}_{\bar{k}}^n,\mu_l)$ and $H^2(\mathbb{P}_{\bar{k}}^n,\mu_l)$, which are $\mu_l,0$ and $\mathbb{Z}/l$ respectively. Thus we have
$$0 \rightarrow H^2(G_k,\mu_l)\rightarrow H^2(\mathbb{P}_k^n,\mu_l) \rightarrow H^0(G_k,\mathbb{Z}/l)\rightarrow 0.$$
Thus $H^2(G_k,\mu_l)\cong \text{coker}(\text{Pic}(\mathbb{P}_k^n)\rightarrow H^2(\mathbb{P}_k^n,\mu_l))[l]\cong Br(\mathbb{P}_k^n)[l].$ Then we note that $H^2(G_k,\mu_l)\cong Br(k)[l]$. Since this is true for all $l$, we see that in fact $Br(k)\cong Br(\mathbb{P}_k^n)$.
