TITLE: A simple proof of the Weyl algebra's rigidity.
QUESTION [10 upvotes]: I am wondering if there is a nice presentation of the Hochschild cohomology of $A_n$ the Weyl algebra. It is known that $H^m(A_n,A_n)=0$ for $m>0$, and thus it is rigid. A proof can be found in Sridharan, but this proof seems to be doing a lot more and is fairly complicated. 
I was wondering if there was a simpler way to see this fact specifically. Essentially, I am being a bit lazy. 
Thanks!

REPLY [11 votes]: 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.