已知数列 \(\left\{ {{a_n}} \right\}\) 的前\(n\)项和为 \({S_n}\) ，且 \({S_n} = 2{a_n} - 1\) ，则数列 \(\left\{ {{a_n}} \right\}\) 的通项公式\({a_n} = \)___．

当 \(n = 1\) 时， \({a_1} = 2{a_1} - 1\) ，解得： \({a_1} = 1\) ；

当 \(n \geqslant 2\) 时， \({a_n} = {S_n} - {S_{n - 1}} = 2{a_n} - 1 - \left( {2{a_{n - 1}} - 1} \right)\) ， \(\therefore {a_n} = 2{a_{n - 1}}\) ，

则数列 \(\left\{ {{a_n}} \right\}\) 是以\(1\)为首项，\(2\)为公比的等比数列， \(\therefore {a_n} = 1 \times {2^{n - 1}} = {2^{n - 1}}\) .

故答案为：\({2^{n - 1}}\).

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