Here is my question:
Let $\alpha_0, \dots, \alpha_{p-1} \in \mathbb{Z}_p$ be such that $\alpha_i \equiv i \pmod{p}$ for all $i = 0,\dots, p-1$. Show that, for any $x\in \mathbb{Z}_{p}$, you can find an infinite sequence $(a_{n})_{n\geq 0}$, where each $a_n \in \lbrace\alpha_0,\alpha_1,\dots,\alpha_{p-1} \rbrace$, such that $$x = \sum_{n=0}^{\infty}a_np^{n}.$$
Any suggestions?
Take the class $\bar x$ of $x\bmod p^n$. Since $\Bbb Z_p/p^n\Bbb Z_p\simeq\Bbb Z/p^n\Bbb Z$, you can find $$ m_n=a_0+a_1p+a_2p^2+...+a_{n-1}p^{n-1}\in\Bbb Z $$ such that $\bar x=\bar m_n$, where each $a_i\in\{0,1,...,p-1\}$.
Now observe that $$ m_{n+1}=m_n+a_np^n $$ so that the sequence $\{m_n\}$ converges to $x=\sum_{k=0}^\infty a_kp^k$ in $\Bbb Z_p$.