Let us prove the somewhat stronger claim:
Theorem: Every anti-chain in $\langle\mathbb N^n,\leq_{cw}\rangle$ is finite.
Before that we will prove two small Lemma's which will make the proof of the above claim much simpler.
Lemma I: Suppose $x\in\mathbb N^n$, then $\{y\in\mathbb N^n\mid y\leq_{cw} x\}$ is finite.
Proof: Note that $y\leq_{cw} x$ if and only if for all $i we have $y(i)\leq x(i)$. This means that there are at most $\max\{x(i)\mid i many possible vectors with this property, as wanted. $\square$
Lemma II: Suppose $x_1,\ldots x_k\in\mathbb N^n$ then there is $y\in\mathbb N^n$ such that for all $i\leq k$, $x_i\leq_{cw} y$.
Proof: Let $y(i)=\max\{x_j(i)\mid j=1,\ldots,k\}$. It is clear that $y$ is as wanted. $\square$
Finally we prove the theorem:
Proof: Suppose $M\subseteq\mathbb N^n$ is an infinite set, we will find two elements comparable. Let $x_i\in M$ be such that the $i$-th coordinate of $x_i$ is minimal in $M$, now let $x\in\mathbb N^n$ be as described in Lemma II.
Since $T=\{y\in\mathbb N^n\mid y\leq_{cw} x\}$ is finite, we have that $M\setminus T$ is infinite therefore it contains some $y$, but $y$ is necessarily larger than all $x_i$'s we have constructed therefore we have at least two elements comparable in $M$ (I am saying at least two because it is possible that the $x_i$'s are all the same). $\square$
Corollary: If $M\subseteq\mathbb N^n$ is non-empty then $\min M$ is finite.
Proof: Using the theorem above it is enough to show that this set forms an anti-chain, which is immediate from the definition of $\min M$. $\square$
(Do note, that the Corollary is equivalent to the Theorem. If we assume the Corollary, take $M$ to be an anti-chain then $M=\min M$, and therefore it is finite.)