We will show that: :$\forall \epsilon \in \R_{>0}: \size {a^{x y} - \paren {a^x}^y} < \epsilon$ {{WLOG}}, suppose that $x < y$. Consider $I := \closedint x y$. Let $I_\Q = I \cap \Q$. Let $M = \max \set {\size x, \size y}$ Fix $\epsilon \in \R_{>0}$. From Real Polynomial Function is Continuous: :$\exists \delta' \in \R_{>0}: \size {a^x - a^{x'} } < \delta' \leadsto \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} } < \dfrac \epsilon 4$ From Power Function on Strictly Positive Base is Continuous: {{begin-eqn}} {{eqn | lll=\exists \delta_1 \in \R_{>0} :       | ll= \size {x x' - y y'}       | lo= <       | l = \delta_1       | o = \leadsto       | m = \size {a^{x x'} - a^{x y'} }       | mo= <       | r = \dfrac \epsilon 4 }} {{eqn | lll=\exists \delta_2 \in \R_{>0} :       | ll= \size {x y' - x' y'}       | lo= <       | l = \delta_2       | o = \leadsto       | m = \size {a^{x y'} - a^{x' y'} }       | mo= <       | r = \dfrac \epsilon 4 }} {{eqn | lll=\exists \delta_3 \in \R_{>0} :       | ll= \size {x' - x}       | lo= <       | l = \delta_3       | o = \leadsto       | m = \size {a^{x'} - a^x}       | mo= <       | r = \delta' }} {{eqn | o = \leadsto       | m = \size {\paren {a^x}^{y'} - \paren {a^{x'} }^{y'} }       | mo = <       | r = \dfrac \epsilon 4 }} {{eqn | lll=\exists \delta_4 \in \R_{>0} :       | ll= \size {y' - y}       | lo= <       | l = \delta_4       | o = \leadsto       | m = \size {\paren {a^x}^{y'} - \paren {a^x}^{y} }       | mo= <       | r = \dfrac \epsilon 4 }} {{end-eqn}} Further: {{begin-eqn}} {{eqn | l = \size {y - y'}       | o = <       | r = \frac {\delta_1} {\size x} }} {{eqn | ll= \leadsto       | l = \size {x y - x y'}       | r = \size x \size {y - y'}       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | o = <       | r = \size x \frac {\delta_1} {\size x}       | c = multiplying both sides by $\size x \ge 0$ }} {{eqn | r = \delta_1 }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = \size {x - x'}       | o = <       | r = \frac {\delta_2} M }} {{eqn | ll= \leadsto       | l = \size {x y' - x'y'}       | r = \size {y'} \size {x - x'}       | c = Absolute Value Function is Completely Multiplicative }} {{eqn | o = \le       | r = M \size {x - x'}       | c = Real Number Ordering is Compatible with Multiplication }} {{eqn | o = <       | r = M \frac {\delta_1} M       | c = multiplying both sides by $M \ge 0$ }} {{eqn | r = \delta_2 }} {{end-eqn}} Let $\delta = \max \set {\dfrac {\delta_1} {\size x}, \dfrac {\delta_2} M, \delta_3, \delta_4}$. From Closure of Rational Interval is Closed Real Interval: :$\exists r, s \in I_\Q: \size {x - r} < \delta \land \size {y - s} < \delta$ Thus: {{begin-eqn}} {{eqn | l = \size {a^{x y} - \paren {a^x}^y}       | o = \le       | r = \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {a^{r s} - \paren {a^r}^s} + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}       | c = Triangle Inequality for Real Numbers }} {{eqn | r = \size {a^{x y} - a^{x s} } + \size {a^{x s} - a^{r s} } + \size {\paren {a^r}^s - \paren {a^x}^s} + \size {\paren {a^x}^s - \paren {a^x}^y}       | c = Product of Indices of Real Number: Rational Numbers }} {{eqn | o = <        | r = \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4 + \frac \epsilon 4       | c = Definition of $r$ and $s$ }} {{eqn | r = \epsilon }} {{end-eqn}} Hence the result, by Real Plus Epsilon. {{qed}}