If $M$ is a finite length module over a noetherian local ring $R$, then is $\mathrm{Ext}_R^i(M, R)$ finite length as well? 
Let $R$ be a noetherian local ring. Is it true that if $M$ is a finite length $R$-module, then for all $k \geq 0$, $\mathrm{Ext}_R^k(M, R)$ is finite length as well?

I can show the $k = 0$ case, since $\mathrm{Hom}_R(M, R)$ is a (finitely generated) ideal of $R$ which is annihilated by a power of the maximal ideal. However, I'm not sure how to boot-strap this up to all $k$.  
 A: Let $(R,\mathfrak{m})$ be a Noetherian local ring. If $M$ is a finite length $R$-module, the only associated prime of $M$ is $\mathfrak{m}$ so there is a filtration $$ 0 = M_{0} \subset \dotsb \subset M_{n} = M $$ of $M$ where each successive quotient $M_{i}/M_{i-1}$ is isomorphic to the residue field $R/\mathfrak{m}$. Using the long exact sequence in Ext, we may reduce to the case where $M = R/\mathfrak{m}$. In this case $\mathrm{Ext}_{R}^{k}(R/\mathfrak{m},R)$ is a finitely generated $R$-module which contains $\mathfrak{m}$ in its annihilator. To see that $\mathrm{Ext}_{R}^{k}(R/\mathfrak{m},R)$ is annihilated by $\mathfrak{m}$, note that one way to compute $\mathrm{Ext}_{R}^{k}(R/\mathfrak{m},R)$ is to take a resolution $R \to I^{\bullet}$ of $R$ by injective $R$-modules, apply $\mathrm{Hom}_{R}(R/\mathfrak{m},-)$ to $I^{\bullet}$, then take cohomology. We have that $\mathrm{Hom}_{R}(R/\mathfrak{m},N)$ is annihilated by $\mathfrak{m}$ for any $R$-module $N$.
A: Over a Noetherian ring $R$, finitely generated modules $M,N$ and $f\in R$, one has $(\mathrm{Ext}^k_R(M,N))_f=\mathrm{Ext}^k_{R_f}(M_f,N_f)$. Can you finish the rest?
