Let $\alpha\in\mathbb{C}$, and let $\mathbb{Z}[\alpha]$ be the smallest subring of $\mathbb{C}$ containing $\alpha$; that is, $\mathbb{Z}[\alpha]=\cap\ S$, where $S$ ranges over all those subrings of $\mathbb{C}$ containing $\alpha$. Prove that $$\mathbb{Z}[\alpha]=\{f(\alpha): f(x)\in\mathbb{Z}[x]\}$$
My attempt:
I think I know how to solve this problem but don't know how to present it as a "mathematical" proof.
Since $\mathbb{Z}[\alpha]$ is a subring of $\mathbb{C}$, it contains $0$ and $1$.
Since $\mathbb{Z}[\alpha]$ is closed under addition, it must contain all the natural numbers. It must also contain the additive inverses of the natural numbers, i.e, the negative integers. Thus $\mathbb{Z}[\alpha]$ contains $\mathbb{Z}$.
Since $\mathbb{Z}[\alpha]$ is closed under multiplication, it must contain all elements of the form $m\alpha^n$ where $m\in\mathbb{Z}$ and $n\in\mathbb{N}\cup\{0\}$.
Finally, since $\mathbb{Z}[\alpha]$ is closed under addition, it must contain all elements of the form $\sum m\alpha^n$ where $m$ and $n$ range over $\mathbb{Z}$ and $\mathbb{N}\cup\{0\}$ respectively.
How do I finish this proof and make it "mathematically presentable"?