Total order on $\mathbb{Z}^{n}_{\geq 0}$ 
Suppose that $ < $ is a total order on $\mathbb{Z}^{n}_{\geq 0}$ satisfying:
1.) $\alpha < \beta \iff \alpha + \gamma < \beta + \gamma$ for all $\alpha ,\beta , \gamma$
2.) $\vec{0}$ is the smallest element in $\mathbb{Z}^{n}_{\geq 0}$ in this order.
Must it follow that $\mathbb{Z}^{n}_{\geq 0}$ is well-ordered under this ordering? That is, every non-empty subset has a least element?

 A: Completely revised.
Let $\alpha=\langle a_1,\ldots,a_n\rangle,\beta=\langle b_1,\ldots,b_n\rangle\in\Bbb Z_{\ge 0}^n$. If $a_k\le b_k$ for $k\in[n]$, then $\vec 0\le\beta-\alpha$, so $\alpha\le\beta$. Thus, $\alpha>\beta$ implies that $a_k>b_k$ for some $k\in[n]$.
Now suppose that there are $\alpha_k=\langle a_1^{(k)},\ldots,a_n^{(k)}\rangle\in\Bbb Z_{\ge 0}^n$ for $k\in\Bbb Z_{\ge 0}$ such that $\alpha_k>\alpha_{k+1}$ for each $k\in\Bbb Z_{\ge 0}$. Let $A=\{\alpha_k:k\in\Bbb Z_{\ge 0}^n\}$, and let $P$ be the set of two-element subsets of $A$. For each $\{\alpha_k,\alpha_\ell\}\in P$ let 
$$c:P\to[n]:\{\alpha_k,\alpha_\ell\}\mapsto\min\left\{i\in[n]:a_i^{(\min\{k,\ell\})}>a_i^{(\max\{k,\ell\})}\right\}\;.$$
The infinite Ramsey theorem ensures that there are an infinite $A_0\subseteq A$ and an $i_0\in[n]$ such that $c(\{\alpha_k,\alpha_\ell\})=i_0$ whenever $\alpha_k$ and $\alpha_\ell$ are distinct elements of $A_0$. But this is impossible, since $\langle a_{i_0}^{(k)}:\alpha_k\in A_0\rangle$ is then a strictly decreasing infinite sequence of non-negative integers. Thus, there is no infinite descending chain in $\langle\Bbb Z_{\ge 0}^n,\le\rangle$, which is therefore a well-order.
