Every Derivation of $M_n (\mathbb{R})$ are inner? How I can prove this proposition?

Proposition: Every Derivation of $M_n (\mathbb{R})$ with respect to matrix multiplication are inner.

 A: Let $K$ be an algebraically closed field of characteristic $0$, $L$ be the Lie algebra $M_n(K),[.,.]$, $ad(L)=\{ad_x:y\rightarrow xy-yx=[x,y]|x\in L\}$, $der(L)$ be the set of derivations over $L$. 
Prop 0. $L$ has no proper ideal. (well-known)
Prop 1. $ad(L)$ is an ideal of $der(L)$. 
Proof. $[\delta,ad_x]=ad_{\delta x}$.
Prop 2. $x\in L\rightarrow ad_x\in ad(L)$ is an isomorphism of Lie algebra.
Proof. $ad_{[x,y]}=[ad_x,ad_y]$.
Prop 3. The Killing form over $L$: $K_L(x,y)=Tr(ad_x.ad_y)$ is non-degenerate and $K([x,y],z)=K(x,[y,z])$.  
Proof. Note that $S=\{x\in L|$ for every $y\in L,K_L(x,y)=0\}$ is an ideal of $L$. It is easy to find $x,y$ s.t. $K(x,y)\not= 0$; according to 0., $S=\{0\}$.
Remark. $K_{ad(L)}$ is the restriction of $K_{der(L)}$ and, according to 2.,3., is non-degenerate. Let $I$ be the orthogonal of $ad(L)$ in $der(L)$ with respect to $K_{der(L)}$; then $I\cap ad(L)=\{0\}$; according to 1., they are ideals of $der(L)$ s.t. $[I,ad(L)]\subset ad(L)\cap I=\{0\}$.
Let $\delta\in I$; according to 1., for every $x$, $ad_{\delta x}=0$; according to 2., for every $x$, $\delta_x=0$ and $\delta= 0$. Conclusion $ad(L)=der(L)$.
A: Matrices are Morita equivalent to their underlying ring, and Morita equivalent rings have isomorphic Hochschild cohomology. A field as an algebra over itself has trivial cohomology for nonzero degrees.
