User:Keith.U/Sandbox

Proof
This proof assumes the Exponent of Sum property.

First, for $n \in \N$:

That is:
 * $\forall n \in \N : \exp \left({ ny }\right) = \left({ \exp y }\right)^{n}$

Next:

The ultimate equality of which holds from Exponential of Zero.

Thus:
 * $\forall m \in \Z : \exp \left({ my }\right) = \left({ \exp y }\right)^{m}$

Further:

So fix $r \in \Q$.

Let $r = \dfrac{m}{n}$, where $m \in \Z, n \in \N$.

From above:

Thus, from the definition of $\left({ \exp y }\right)^{x}$ as the unique continuous extension of $r \mapsto \left({ \exp y }\right)^{r}$ from $\Q$ to $\R$:
 * $\exp \left({ xy }\right) = \left({ \exp y }\right)^{x}$