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?

Yes, for $N=R$ the result is true.

Let $R$ be a noetherian ring and $M$ a finitely generated $R$-module with $\mathrm{pd}_R(M)=n$. Then $\mathrm{Ext}_R^n(M,R)\neq 0$. 

Proof. First consider $R$ local with maximal ideal $\mathfrak{m}$. Let $F_{\bullet}$ be a minimal free resolution of $M$ and $\varphi_n:F_n\rightarrow F_{n-1}$ the last homomorphism. One knows that the elements of the matrix associated to $\varphi_n$ are all in $\mathfrak{m}$. $\mathrm{Ext}_R^n(M,R)$ is the cokernel of the dual map $\varphi_n^*:F_{n-1}^*\rightarrow F_n^*$ whose matrix is the transpose of the matrix associated to $\varphi_n$. Therefore the elements of the matrix associated to $\varphi_n^*$ are also in $\mathfrak{m}$, and this shows that $\mathrm{Coker}\ \varphi_n^*\neq 0$.
Now it is easy to deduce the general case from the local case via the well known fact that $\mathrm{pd}_R(M)= \sup_{\mathfrak{m}\in\mathrm{Max}(R)}\mathrm{pd}_{R_{\mathfrak{m}}}(M_{\mathfrak{m}})$.
