Let $X$ be a (EDIT: non-singular, or even $\mathbf A^n$) algebraic variety over a field $k$ (alg. closed). Suppose $G$ is a finite group acting on $X$, $|G|\neq 0$ in $k$. Then $k[X]^G$ is Cohen-Macaulay. All the proofs I've seen are fairly involved.

- Question: is there a simple proof of this fact?