There are infinitely many monomial orders 
Show that if $n ≥ 2$ there are infinitely many monomial orders on  $k[x_1, \ldots , x_n]$.

I think it is Robbiano theorem (with the exception $n>2$) at the link below but i can't understand proof: why we take $A,B$ from $\mathbb N^n $? 
http://www.ces.clemson.edu/~janoski/reu/2012/REULec2Mohammed.pdf
If you have different proof other than link please say it or help about proof at link. 
 A: A monomial $x_1^{a_1}\cdots x_n^{a_n}$ can be identified with the vector $A=(a_1,\dots,a_n)$ from $\mathbb N^n$.
Let's prove that theorem for $n=2$ which is actually enough (why?). We know that $a_1+a_2\gamma>0$ iff $a_1+a_2\tau>0$, where $a_1,a_2\in\mathbb Z$, and want to prove that $\gamma=\tau$. If, let's say, $\gamma >\tau$, then take a rational number $q=-a/b$ (with $b>0$) such that $\gamma>q>\tau$. Thus we have $a+b\gamma>0>a+b\tau$, a contradiction. 
A: A monomial order on $k[x_1,\dots,x_n]$ is the same as an order on $\mathbb N^n$, because we can identify a monomial $x_1^{a_1}\cdots x_n^{a_n}$ with the vector $(a_1,\dots,a_n) \in \mathbb N^n$.
The rest of the proof is not very clearly written. Here's the beginning of an explanation. To show that $\geq_u \neq \geq_v$, we have to see that there exists $A,B \in \mathbb N^n$ with $A \geq_u B$ but with $B \geq_v A$. Clearly, this is equivalent to showing that there exists $C \in \mathbb Z^n$ with $C \cdot u \geq 0$, but $C \cdot v \leq 0$, where, as in the proof, $u=(1,\gamma,\gamma^2,\dots,\gamma^{n-1})$ and $v=(1,\tau,\dots,\tau^{n-1})$.
If this were not true, then the sign of $C \cdot u$  and $C \cdot v$ would be the same for all $C$. But this is equivalent to saying that $(C \cdot u)(C \cdot v) > 0$ for all $C$.
But if $(C \cdot u)(C \cdot v)=0$, then we would have a relation of the form
$$ c_0^2 + c_0c_1(\gamma + \tau) + c_0c_2(\gamma^2+\tau^2)+c_1^2(\gamma \tau)+\cdots = 0 $$
But this shouldn't be possible for different transcendental $\gamma,\tau$, i.e. they are linearly independent over $\mathbb Q$.
