How many elements does $\mathbb{Z}_{p^{\alpha}}^*$ have? 

Let $p\in\mathbb{N}, p>2$ be a prime number and $\alpha\in\mathbb{N},\alpha\geqslant 2$.
    How many elements does $\mathbb{Z}_{p^{\alpha}}^*$ have?


What I know is that
$$
\mathbb{Z}_{p^{\alpha}}^*=\left\{x+p^{\alpha}\mathbb{Z}| x\in\left\{0,\ldots,p^{\alpha}-1\right\}: gcd(x,p^{\alpha})=1\right\}\\=\left\{x+p^{\alpha}\mathbb{Z}| x\in\left\{0,\ldots,p^{\alpha}-1\right\}: gcd(x,p)=1\right\}
$$
so the remaining question (which is probably a silly one, but I do not know) is for how many $x\in\left\{0,\ldots,p^{\alpha}-1\right\}$ it is $gcd(x,p)=1$.
How can I answer this?

You are almost there. See some easy concrete examples first (e.g. $p^\alpha=8$ or $9$ or $27$).
You will recognise that every $p$th number is a multiple of $p$, but no more. So, until $x=p^\alpha$ we have exactly $\displaystyle\frac{p^\alpha}p$ numbers $x$ that are divisible by $p$, and the rest will have $\gcd(x,p)=1$.
