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I'm trying to show that the function $f_n :[0,\infty )\rightarrow [0,\infty ) $ given by $x\mapsto x^n$ satisfies the property $x^{\alpha + \alpha ' }=x^{\alpha } x^{\alpha ' } , \alpha , \alpha ' \in \mathbb{Q} .$

I'm supposed to use that $x\mapsto x^{nn'} $ is injective but I'm unsure on how to show this.

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Are you asking why $x^{\alpha+\alpha'}= x^\alpha x^{\alpha'}$ for any pair of rational points $\alpha = \frac{a}{b}, \alpha' = \frac{a'}{b'}$? If this is what you mean, try showing that $(x^{\alpha + \alpha'})^{bb'} = \left ((x^\alpha)(x^{\alpha'})\right)^{bb'}$. Injectivity of the map $t \mapsto t^{bb'}$ will then give you the desired equality $x^{\alpha+\alpha'}= x^\alpha x^{\alpha'}$.

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    For RHS I get $x^{a'b}.x^{ab'} $ and for LHS I get $ x^{a'b+ab'} $ but I don't get how I can prove that these are equal. – 2017-02-16
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    $a'b$ and $ab'$ are both integers. It is a consequence of definition that $x^{p + q} = x^px^q$ whenever $p,q$ are integers. – 2017-02-16