 Solution

Suppose all functions $\mathbb{N}\to \mathbb{N}$ could be listed as $f_1,f_2,\dots$. DefineThen $g$ differs from $f_n$ at $n$ for every $n$, so it is absent from the list. This Cantor's diagonal argument proves that ${\mathbb{N}}^{\mathbb{N}}$ is an uncountable set.