Adjoint semi-simple algebraic groups over non-algebraically closed fields Let $k$ be a field of characteristic zero and let $G$ be an adjoint semi-simple algebraic group over $k$.
On p34 of the paper "Sansuc - Groupe de Brauer et arithmétique des groupes
algébriques lineaires sur um corps de nombres", it is claimed that there exists a collection of finite field extensions $k \subset k_i$ such that
$$G \cong \prod_i \mathrm{R}_{k_i/k}(G_i) \quad (*)$$
where the $G_i$ are absolutely simple adjoint groups over $k_i$ and $\mathrm{R}_{k_i/k}$ denotes the Weil restriction.
I was quite surprised when I saw this, as it certainly seems to be something special about adjoint groups. However, Sansuc unfortunately gives no explanation nor reference why this holds.

Why does the stated isomorphism (*) exist?

I would be happy with either a proof or a reference.
 A: See Proposition 6.4.4 and Remark 6.4.5 in Brian Conrad's article "Reductive groups schemes" in "Autour des schemas en groupes, Vol. I" (alternatively, http://math.stanford.edu/~conrad/papers/luminysga3smf.pdf) for a proof (in a more general setting). 
As you mention, this is special for adjoint groups (but would also work for simply connected groups).
A: The proof is quite easy. Let $G$ be a simply connected (or adjoint) semisimple
group over a field $k$. Over the separable closure $k^{s}$ of $k$, $G$ is a
product $G=G_{1}\times\cdots\times G_{n}$ of almost-simple groups $G_{i}$. The
Galois group $\Gamma$ of $k^{s}/k$ acts on the set $\{G_{1},\ldots,G_{n}\}$
and the product of the groups in an orbit is stable under $\Gamma$, and hence
defined over $k$. In this way, $G$ is a product of quasi-simple groups over
$k$. Thus, we may suppose that $G$ itself is quasi-simple. Now $\Gamma$ acts
transitively on the set  $\{G_{1},\ldots,G_{n}\}$. Let $\Delta$ be the
stabilizer of $G_{1}$, and let $K$ be the subfield of $k^{s}$ fixed by
$\Delta$. Then $Res_{K/k}(G_{1})$ and $G$ are isomorphic over $k^s$, by an
isomorphism invariant under the action of of $\Gamma$, and so they are
isomorphic over $k$.
