We can easily seen that $${q^{2n}} = {q^n}\left( {{q^n} - 1} \right) + {q^n},{q^{3n}} = {q^n}{\left( {{q^n} - 1} \right)^2} + 2{q^n}\left( {{q^n} - 1} \right) + {q^n}.$$ Similarly, we assume that $${q^{mn}} = {q^n}\sum\limits_{k = 1}^{m - 1} {{a_k}{{\left( {{q^n} - 1} \right)}^k}} + {q^n}.$$ then, how to obtain the coefficient $a_k$.
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How about taking $q^{mn}$ as $q^n((q^n-1)+1)^{m-1}$, then use binomial theorem? – Asydot Sep 02 '15 at 05:23
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1That's the answer....Why not post it as an answer? – DanielWainfleet Sep 02 '15 at 08:00