Identity Theorem Let $(\alpha_n)$ be a sequence of complex numbers such that $$ \sum_{n=1}^\infty \frac{\alpha_n}{k^n} = 0 $$ for all $k = 1, 2, 3, ...$ Prove that $\alpha_n=0$ for all $n$
As far as a solution goes, really not sure where to start. Any hints would be appreciated. Thanks.

Following Xavier's excellent hint:
$$\alpha\neq 0\;\;,\;\;k\in\Bbb N\Longrightarrow \sum_{n=0}^\infty\frac{\alpha^n}{k^n}=\sum_{n=0}^\infty\left(\frac{\alpha}{k}\right)^n=\begin{cases}\text{diverges}&\text{if}\,\;\;\left|\frac{\alpha}{k}\right|\geq 1\\{}\\\frac{1}{1-\frac{\alpha}{k}}=\frac{k}{k-\alpha}&\text{if}\,\;\;\left|\frac{\alpha}{k}\right|<1\end{cases}\;\;\neq 0$$
