= Solution
Suppose, for a contradiction, that the monic <irreducible polynomials> in the <polynomial ring> $k[t]$ form a finite list $p_1,\ldots,p_r$. The list is nonempty because $t$ itself is monic and irreducible. Put
$$
F(t)=p_1(t)\cdots p_r(t)+1.
$$
This has positive <degree>, so it is a nonzero <nonunit>. Every such <polynomial> over a <field> has an <irreducible polynomial> as a factor: repeatedly factor a reducible polynomial, with the positive degrees strictly decreasing until an irreducible factor is reached. Dividing that factor by its <leading coefficient> makes it monic; call it $q$.
By the assumed completeness of the list, $q=p_j$ for some $j$. But then $p_j$ divides both $F$ and $p_1\cdots p_r$, hence their difference $1$. A positive-degree <polynomial> cannot divide $1$. This contradiction proves \b[there are infinitely many monic irreducible polynomials over every field]. This is the polynomial version of <Euclid's theorem>; it does not require the field itself to be infinite.
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