Any substructure of $(\mathbb{N}; 0, 1, +, \cdot)$ is itself

Question

Consider a substructure $\mathcal{M} \subseteq \mathcal{N} = (\mathbb{N}; 0, 1, +, \cdot)$. Prove that $\mathcal{M} = \mathcal{N}$.

EDIT: This result seems intuitively easy, but I'm having trouble finding a formal proof in first-order logic (i.e. by using the language of definable sets, Tarski's criterion, substructures, and so forth) which I'd like to see or get a hint for.

Answer 1

$0,1\in\mathcal{M}$; close this set under $+$.