The result can be proved by induction on $n$. It's certainly true for $n=1$. If it's not true for all $n$, let $m$ be minimal such that it's not true for $m$. Then there is some $M\subseteq\Bbb N_0^m$ that has an infinite set $\{\langle a_k(1),\dots,a_k(m)\rangle:k\in\Bbb N_0\}$ of minimal elements.
Suppose that the sequence $\langle a_k(1):k\in\Bbb N_0\rangle$ has an infinite constant subsequence $\langle a_{k(i)}(1):i\in\Bbb N_0\rangle$. Then $\{\langle a_{k(i)}(2),\dots,a_{k(i)}(n)\rangle:i\in\Bbb N_0\}$ is an subset of $\Bbb N_0^{m-1}$, all of whose elements are minimal, contradicting the choice of $m$. It follows easily that $\langle a_k(1):k\in\Bbb N_0\rangle$ has a strictly increasing infinite subsequence $\langle a_{k(i)}(1):i\in\Bbb N_0\rangle$. Thus, we can replace $\{\langle a_k(1),\dots,a_k(m)\rangle:k\in\Bbb N_0\}$ by its infinite subset $\{\langle a_{k(i)}(1),\dots,a_{k(i)}(m)\rangle:i\in\Bbb N_0\}$, and we might therefore just as well assume that $\langle a_k(1):k\in\Bbb N_0\rangle$ is strictly increasing to begin with.
By repeating this argument on the other coordinates, we may assume that $\langle a_k(i):k\in\Bbb N_0\rangle$ is strictly increasing for $i=1,\dots,m$. But then $$\big\langle a_k(1),\dots,a_k(m)\big\rangle <_{cw}\big\langle a_{k+1}(1),\dots,a_{k+1}(m)\big\rangle$$ for every $k\in\Bbb N_0$, which very badly violates the choice of $\{\langle a_k(1),\dots,a_k(m)\rangle:k\in\Bbb N_0\}$. This contradiction establishes the desired result.
Added: I finally managed to track down a vagrant memory: this result is sometimes known as Dickson's lemma. It's a special case of the more general result that if $\langle Q,\le\rangle$ is a well-quasi-order, then so is $\langle Q^n,\le_{cw}\rangle$ for any positive integer $n$.