(My $\Bbb N$ is your $\Bbb Z_{\ge 0}$.) Suppose that $\left\langle a^{(k)}:k\in\Bbb N\right\rangle$ is a sequence in $\Bbb N^n$ such that $a^{(k+1)}<_g a^{(k)}$ for each $k\in\Bbb N$, where the subscript $g$ denotes the grevlex order. Let $a^{(k)}=\left\langle a_1^{(k)},\ldots,a_n^{(k)}\right\rangle$ for each $k\in\Bbb N$. For $a=\langle a_1,\ldots,a_n\rangle\in\Bbb N^n$ let $\deg a=a_1+\ldots+a_n$. Then $\left\langle\deg a^{(k)}:k\in\Bbb N\right\rangle$ is a non-increasing sequence of non-negative integers, so it must be eventually constant: there are $k_0,m\in\Bbb N$ such that $\deg a^{(k)}=m$ for all $k\ge k_0$. Without loss of generality we may as well assume that $k_0=0$ and just look at the tail of the sequence, since all we want to show is that the sequence actually has to be finite.
Now for $k\ge 1$ let $\ell(k)$ be the largest subscript such that $a_{\ell(k)}^{(k)}\ne a_{\ell(k)}^{(0)}$; it’s not hard to check that the sequence $\langle\ell(k):k\ge 1\rangle$ is non-decreasing, so there are $k_1\in\Bbb Z^+$ and $\ell\in\{1,\ldots,n\}$ such that $\ell(k)=\ell$ for all $k\ge k_1$. Once again there is no harm in assuming that $k_1=1$, so that $\ell(k)=\ell$ for all $k\ge 1$.
This means that the sequence $\left\langle a_\ell^{(k)}:k\in\Bbb N\right\rangle$ is strictly increasing. But $\deg a^{(k)}=m$ for all $k\in\Bbb N$, and $$\sum_{i=\ell+1}^na_i^{(k)}=\sum_{i=\ell+1}^na_i^{(0)}$$ for all $k\in\Bbb N$, so
$$\left\langle\sum_{i=1}^{\ell-1}a_i^{(k)}:k\in\Bbb N\right\rangle\tag{1}$$
must be strictly decreasing. But this is impossible, since $(1)$ is a sequence of non-negative integers. Thus, $\langle\Bbb N^n,\le_g\rangle$ has no infinite strictly decreasing sequence.
The key idea here is that if you hold fixed the degree of a monomial in $n$ indeterminates, there are only finitely many possible combinations of exponents.
