Modules with projective dimension $n$ have not vanishing $\mathrm{Ext}^n$ 
Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module with projective dimension $n$. Then for every finitely generated $R$-module $N$ we have $\mathrm{Ext}^n(M,N)\neq 0$. Why?

By definition, if the projective dimension is $n$ this means that $\mathrm{Ext}^{n+1}(M,-)=0$ and $\mathrm{Ext}^n(M,-)\neq 0$, so there exists an $N$ such that $\mathrm{Ext}^n(M,N)\neq 0$. Why is this true for every $N$?
I found this theorem on this notes, page 6 proposition 9. Did I misunderstand it? Is there a similar statement?
This result is used on page 41, implication 3 implies 4, and is used in the case $N=R$. Is it true in this case?

Suppose $R$ is local noetherian with maximal ideal $\def\m{\mathfrak m}\m$ and residue field $k$, and let $M$ be a f.g. $R$-module with $\operatorname{pdim}M=n$. Then $\def\Ext{\operatorname{Ext}}\Ext_R^{n}(M,k)\neq0$.
Let $N$ be a non-zero f.g. module. There is a submodule $N'\subseteq N$ such that $N/N'\cong k$ as $R$-modules. From the short exact sequence $0\to N'\to N\to k\to 0$ we get an exact sequence $$\Ext^n_R(M,N)\to\Ext^n_R(M,k)\to\Ext^{n+1}_R(M,N')=0$$ Since the middle term is not zero and the leftmost map is surjective, we see that $\Ext^n_R(M,N)\neq0$.
Remove now the hypothesis that $R$ be local and let again $M$ be finitely generated and of projective dimension $n$. There is a maximal ideal $\def\m{\mathfrak m}$ in $R$ such that $\def\pdim{\operatorname{pdim}}\pdim_RM=\pdim_{R_\m}M_\m$.
Then $\def\Ext{\operatorname{Ext}}\Ext^n_R(M,R)_\m\cong\Ext^n_{R_\m}(M_m,R_m)\neq0$, by the local case.
