Let $c_1,\ldots,c_n\in\mathbb R\setminus\{0\}$ and $t_1,\ldots,t_n\in(0,1)$. Is it possible that $\forall k\in\mathbb Z_{>0},\sum_{i=1}^nc_it_i^k=0$? While trying to solve an analysis homework problem, I reduced it to this one.
Let $c_1,\ldots,c_n\in\mathbb R\setminus\{0\}$ and $t_1,\ldots,t_n\in(0,1)$, where $t_i$ are distinct. Is it possible that $\forall k\in\mathbb Z_{>0},\sum_{i=1}^nc_it_i^k=0$?
My thoughts are no, but my algebra is pretty insufficient. I considered $\sum_{i=1}^nc_it_i(1+t_i)=\sum_{i=1}^nc_it_i$ (which follows from assuming the proposition) to try to reach contradiction, but my algebra skills stand in the way of getting anywhere.

This is not possible:
Since the $t_i$ are distinct, we can assume $0 < t_1 < \dots < t_n < 1$. Now assume that $\sum_i c_i t_i^k = 0$ for all $k$. By factoring out the factor $t_n^k$, this implies
$$
0 =  \sum_{i=1}^{n-1} c_i (t_i / t_n)^k + c_n \xrightarrow[k\to\infty]{} c_n,
$$
a contradiction, since $c_n \neq 0$ by assumption. This argument uses that $0 < t_i / t_n < 1$ for $i =1,\dots,n-1$, so that $(t_i/t_n)^k \to 0$ as $k \to \infty$.
