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?

Take $R=\mathbb Z\times\mathbb Z$, consider the elements $e_1=(1,0)$, $e_2=(0,1)\in R$, and the modules  $M=R/(2e_1)$ and $N=Re_2$. Show that the projective dimension of $M$ is $1$ and compute $\operatorname{Ext}_R^1(M,N)$.
