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I'm sure there is an easy argument for the following identity, but I'm unable to figure it out. For $q \in \mathbb{N}$ and $a>0$, why is it that $$ q\ a^{1-1/q} = \sum_{i=0}^{q-1}(a^{1/q})^i(a^{1/q})^{q-1-i}. $$

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up vote 4 down vote accepted

Note that $$(a^{1/q})^i (a^{1/q})^{q-1-i} = (a^{1/q})^{q-1}$$ Therefore, the expression inside the sum is not even dependent on i! And the sum is just this constant, times $q$ (the amount of times it's added).

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