Show that $f(n, k) = \binom{n}{k}$ is an onto function

Question

I have to rigorously prove the statement:

Let $f(n, k) = \binom{n}{k}$, where $n, k \in \mathbb{N}$. Then $f: \mathbb{N} \times \mathbb{N}$ is an onto function.

First I translated the above statement into predicate logic:

$\forall n, k \in \mathbb{N}, f(n,k) = \binom{n}{k} \implies (\forall m \in \mathbb{Z}^{+}, \exists x, y \in \mathbb{N}, f(x, y) = m)$.

How should I approach this proof?

Answer 1

Because $$\binom{n}{1} = n$$ for all $\mathbb N$, it follows that $f$ is onto: for every $n \in \mathbb N$, the pair $(n, 1)$ is a preimage of $n$ under $f$.