A simple proof of the Weyl algebra's rigidity.

(I write $HH^\bullet(\Lambda)$ what you write $H^\bullet(\Lambda,\Lambda)$. When I become Emperor of Notation, everyone will!)

Since $A_n\cong A_1\otimes\cdots\otimes A_1$, the Künneth formula for Hochschild cohomology (which is proved in Cartan-Eilenberg, Theorem XI.3.1, for example) tells you that $HH^\bullet(A_n)\cong HH^\bullet(A_1)^{\otimes n}$. It is enough, then, to compute $HH^\bullet(A_1)$.

The computation of the whole of $HH^\bullet(A_1)$ is not difficult to carry out directly.

You can climb on the shoulders of others and do the following, too. First, it is easy to check that the center of $A_1$ is $k$, so that $HH^0(A_1)=k$. Second, Jacques Dixmier shows, in his Algèbres enveloppentes, that every derivation of $A_1$ is inner, so that $HH^1(A_1)=0$. Finally, the algebra $A_1$ is a Calabi-Yau algebra of global dimension $2$, so in particular it satisfied van den Bergh duality and $HH^2(A_1)=HH_0(A_1)$. The latter vector space is $A_1/[A_1,A_1]$, and a pleasurable computation shows this is zero.

Alternatively, in view of Calabi-Yau-ness of $A_n$, now of global dimension $2n$, we have that $HH^\bullet(A_n)=HH_{2n-\bullet}(A_n)$, and a theorem of Wodzicki tells us that $HH_{2n-\bullet}(A_n)$ is isomorphic to the algebraic de Rham cohomology of $n$-dimensional affine space $\mathbb A^n$, which is just $k$. This way we get $HH^\bullet(A_n)=k$ in one big swoop.

Since the Weyl algebra is a deformation of the polynomial ring in two variables, there is a short transparent proof using deformation theory, cf. M. Gerstenhaber and A. Giaquinto, On the cohomology of the Weyl algebra, the quantum plane, and the q-Weyl algebra, arXiv:1208.0346.