Any quaternion Kählerian manifold of dimension $\geq 8$ is an Einstein space? I wan't to understand the proof this theorem:
Any quaternion Kählerian manifold of dimension $\geq 8$ is an Einstein space.
In the book "Structures on manifolds" by Kentaro Yano and Masahiro Kon is said that this fact is a corollary of these $2$ lemmas.

Lemma $1.$ For any quaternion Kählerian manifold $M$ of dimension
  $n \geq 8$, the Ricci tensor $S$ of $M$ is parallel.
Lemma $2.$  Let $(M,g,V)$ be a quaternion Kählerian manifold of
  dimension $\geq 8$. Then the Riemannian manifold $(M,g)$ is irreducible if
  $(M,g)$ has non-vanishing Ricci tensor.

However I can't see directly how to use them to prove the theorem using the lemmas, neither is it written in the book. Can someone please help me with that. Or tell me where can I read more detailed proof of the theorem? Thanks in advance!
 A: If a manifold is Einstein, then there is a constant $\lambda$ such that $\operatorname{Ric} = \lambda g$. As $g$ is parallel with respect to the Levi-Civita connection, the same is true of the Ricci tensor of an Einstein manifold. What about the converse: if $(M, g)$ has parallel Ricci tensor, is it Einstein? No. If $(M_1, g_1)$ and $(M_2, g_2)$ are two Einstein manifolds with different Einstein constants, then $(M_1\times M_2, g_1 + g_2)$ has parallel Ricci tensor but is not Einstein. In some sense, this is the only possibility, at least locally.

Theorem: Let $(M, g)$ be a Riemannian manifold with parallel Ricci tensor. Then $(M, g)$ is locally isometric to a product of Einstein manifolds.

Returning to the case of a quaternionic Kähler manifold $(M, g)$ of dimension at least $8$, the first lemma tells us that either the Ricci tensor is identically zero or nowhere vanishing. If $\operatorname{Ric} = 0$, then $(M, g)$ is Einstein with $\lambda = 0$. If the Ricci tensor is nowhere vanishing, then the second lemma tells us that $(M, g)$ is irreducible. As the Ricci tensor is parallel, $(M, g)$ is locally isometric to a product of Einstein manifolds, but $(M, g)$ is irreducible, so the product cannot have multiple factors. Therefore $(M, g)$ is locally isometric to an Einstein manifold, so $(M, g)$ is Einstein. That is,

Theorem: Let $(M, g)$ be an irreducible Riemannian manifold with parallel Ricci tensor. Then $(M, g)$ is Einstein.

For a discussion of the two highlighted theorems, see chapter $8$, section $3.1$ of Petersen's Riemannian Geometry, in particular theorems $56$ and $57$. The first highlighted theorem above is not stated in the book, but one can construct a proof by replicating the proof of Theorem $57$ (which is the second highlighted theorem) in the context of Theorem $56$.
If you would like another reference, the theorem you mention is proved in chapter $14$, section $D$ of Besse's Einstein Manifolds, namely Theorem $14.39$ for which two proofs are given. The two lemmas you listed appear as parts $(a)$ and $(b)$ of Theorem $14.45$. Neither of the two proofs use Theorem $14.45$.
