Isomorphism onto $\mathbb{F}_p^{\ast}$, p-adic integers

Question

I want to show, that $\mathbb{Z}_p^\ast/(1+p\mathbb{Z}_p)\cong\mathbb{F}^\ast_p$.

It is $1+p\mathbb{Z}_p\stackrel{?}{=}(p\mathbb{Z}_p)^{\ast}$. So we get $\mathbb{Z}_p^\ast/(p\mathbb{Z}_p)^\ast$.

Now I would like to give an isomorphism $f\colon\mathbb{Z}_p^\ast/(p\mathbb{Z}_p)^\ast\to\mathbb{F}^\ast_p$

But I do not see how I can get such an isomorphism. Is there an easy argument to see this?

Thanks in advance.

Edit: Following the answer of Qiaochu Yuan, I want to show that

$f:\mathbb{Z}_p^\times\to \mathbb{F}_p^\times$ given by $\underbrace{(\dotso, x_n,\dotso, x_1)}_{=x\in\mathbb{Z}_p^\times}\mapsto x_1$

is an isomorphism with kernel $1+p\mathbb{Z}$.

1) $\ker(f)=1+p\mathbb{Z}_p$

Let $x\in\ker(f)$, then $f(x)=x_1\equiv 1\mod p$. Therefore $x=1+x_1p+x_2p^2+\dotso\in 1+p\mathbb{Z}_p$

Let $x\in 1+p\mathbb{Z}_p$, then $x=1+pa$, with $a\in\mathbb{Z}_p$, then $f(x)=1$, hence $x\in\ker(f)$

2) $f$ is surjective:

Let $z\in\mathbb{F}_p^\times$, hence $z\not\equiv 0\mod p$. Write z=z'+pa with $a\in\mathbb{Z}_p$, then $f(z')=z$. And therefore $f$ is surjective.

Answer 1

Any homomorphism of rings $f : R \to S$ gives rise to a homomorphism $R^{\times} \to S^{\times}$ between their unit groups. In particular, the quotient map $f : \mathbb{Z}_p \to \mathbb{F}_p$ (with kernel $p \mathbb{Z}_p$) gives rise to a homomorphism

$$\mathbb{Z}_p^{\times} \to \mathbb{F}_p^{\times}.$$

This homomorphism takes a unit $a_0 + a_1 p + \dots \in \mathbb{Z}_p^{\times}$ and sends it to $a_0 \bmod p$. Now the problem reduces to showing that

1. this homomorphism is surjective, and
2. its kernel is $1 + p \mathbb{Z}_p$.

A nice way to show that this homomorphism is surjective is to show that in fact it has a right inverse sending an element in $\mathbb{F}_p^{\times}$ to its Teichmüller representative, but you don't need to do it this way. Computing the kernel is pretty straightforward.