i think the maximum is $1$ \begin{align*} & \sum\limits_{i=1}^{n}{{{c}_{i}}a_{i}^{k}}=a_{n}^{k}\sum\limits_{i=1}^{n}{{{c}_{i}}}+\sum\limits_{i=1}^{n-1}{\left( \sum\limits_{k=1}^{i}{{{c}_{i}}} \right)}\left( a_{i}^{k}-a_{i+1}^{k} \right) \\ & \le a_{n}^{k}{{n}^{k}}+\sum\limits_{i=1}^{n-1}{{{i}^{k}}}\left( a_{i}^{k}-a_{i+1}^{k} \right) \\ & =\sum\limits_{i=1}^{n}{\left[ {{i}^{k}}-{{\left( i-1 \right)}^{k}} \right]}a_{i}^{k} \\ \end{align*} note that \begin{align*} & {{\left( {{a}_{1}}+{{a}_{2}} \right)}^{k}}\ge a_{1}^{k}+\left( {{2}^{k}}-1 \right)a_{2}^{k} \\ & {{\left( {{a}_{1}}+{{a}_{2}}+{{a}_{3}} \right)}^{k}}\ge {{\left( {{a}_{1}}+{{a}_{2}} \right)}^{k}}+\left( {{3}^{k}}-{{2}^{k}} \right)a_{3}^{k} \\ & \ \cdots \cdots \\ & 1={{\left( \sum\limits_{i=1}^{n}{{{a}_{i}}} \right)}^{k}}\ge {{\left( \sum\limits_{i=1}^{n-1}{{{a}_{i}}} \right)}^{k}}+\left[ {{n}^{k}}-{{\left( n-1 \right)}^{k}} \right]a_{n}^{k} \\ \end{align*} then we are done it is obvious that equality holds when \[\sum\limits_{i=1}^{s}{{{c}_{i}}}={{s}^{k}},{{a}_{1}}={{a}_{2}}=\cdots ={{a}_{s}}=\frac{1}{s},{{a}_{s+1}}=\cdots ={{a}_{n}}=0,s\in \mathbb{Z},1\le s\le n\]