In the proof of Theorem 20.18 in Eisenbud Commutative Algebra, the following fact is stated:
If $S=k[X_1,\ldots,X_r]$ and $N$ is a finite length graded $S$-module, then
$$\operatorname{Ext}_S^r(N,S(-r)) \cong \operatorname{Hom}_k(N,k).$$
It says this follows from Exercise 2.4, but I don't see why...
(I think one needs the fact that if $N$ is of finite length, then $\operatorname{Ext}^j(N,S)=0, \,\, \forall j