Cardinality of the set of functions over the naturals such that $f(n) \geq n$

Question

What is the cardinality of the set of functions over the naturals such that $f(n) \geq n$ for every $n$?

I know it's not countable and at most $\aleph$ so it should be $\aleph$, but I can't find how to bound it from below using a function. Can anyone give me a hint?

Answer 1

Consider the set of functions $g(n) = f(n) - n$, that is just the set of all functions from $\mathbb{N} \to \mathbb{N}$, and this is clearly a biyection.