Well-known Theorem:
Grothendieck’s Finiteness Theorem. Assume that $R$ is a homomorphic image of a regular (commutative Noetherian) ring. Let $\mathfrak a$ be an ideal of $R$, and let $M$ be a finitely generated $R$-module. Then $\lambda_{\mathfrak a}^{\mathfrak a}(M)=f_{\mathfrak a}(M)$ where $$\lambda_{\mathfrak a}^{\mathfrak a}(M)=\inf\{\operatorname{depth}(M_{\mathfrak p})+\operatorname{ht}(\mathfrak a+\mathfrak p)/\mathfrak p : \mathfrak p\in\operatorname{Spec}(R)\setminus \operatorname{Var}(\mathfrak a)\}$$ and $$f_{\mathfrak a}(M)=\inf\{i\in\Bbb N : H^i_{\mathfrak a}(M)\text{ is not finitely generated}\}.$$
My question: Assume that the local ring $(R,\mathfrak m)$ is a homomorphic image of a regular local ring, and let $M$ be a finitely generated $R$-module. I'm looking for a simple proof to show that $f_{\mathfrak m}(M) = \inf\{\operatorname{depth} M_{\mathfrak p} + 1 : \mathfrak p\in \operatorname{Spec}(R)~ \text{and} \dim R/\mathfrak p = 1\}$.
Is the following proof correct?
We have $$f_{\mathfrak m}(M)=\lambda_{\mathfrak m}^{\mathfrak m}(M)=\inf\{\operatorname{depth}(M_{\mathfrak p})+\dim R/{\mathfrak p}: \mathfrak p\in\operatorname{Spec}(R)\setminus \{\mathfrak m\}\}.$$ Let $\operatorname{ht} {\mathfrak m}/{\mathfrak p}=r>1$ and $\mathfrak p= \mathfrak p_0\subset \mathfrak p_1\subset \cdots \subset \mathfrak p_r=\mathfrak m$ then $\operatorname{depth} M \leq \operatorname{depth} M_{\mathfrak p_{r-1}} +1\leq \cdots\leq \operatorname{depth} M_{\mathfrak p} +r=\operatorname{depth} M_{\mathfrak p}+\dim R/{\mathfrak p}$. Therefore conclusion follows.