I keep pointing out to people that how you prove something like this depends heavily upon how you define the terms. And there may be sevaral different ways of defining the same thing.
It is, as you said, easy to show that, for n and m positive integers, a^xa^y= a^{x+y} and (a^x)^y= a^{xy}, for a any positive number, in particular, e. One method of proceeding is to then define a^x, for x not a positive integer, so that those formulas are true. For example, if y= 0, a^xa^y= a^{x+ y} becomes a^xa^0= a^{x+ 0}= a^x. From that, for a any non-zero number, it follows that defining a^0= 1, we are "preserving" that identity. Similarly, any negative integer can be written as -n for n a positive integer. With x= n, y= -n, we have a^na^{-n}= a^{n-n}= a^0= 1 so, again, for a non-zero, we must have a^{-n}= 1/a^n and so we can define a to a negative integer power, a^{-n}, to be the reciprocal of [/itex]a^n[/itex] in order to keep that useful formula true.
If n is a non-zero integer, then (a^{1/n})^{n}= a^{n(1/n)}= a^1= a. That is, in order that that "law of exponents" be true even for non-integer powers, we must have a^{1/n} equal to a number whose nth power is a. In order to be sure such a thing exists (I am assuming real numbers here) we must require that a be positive. In that case, there may be two such real numbers. We define a^{1/n} to be the positive such root. Of course, it follows that a^{m/n}= (a^{1/n})^m so we use that to define a^r for any rational number.
Finally, we define a to an irrational power by requiring that a^x (again, for a any positive number and so, in particular, e^x) be continuous. That is, if \{r_1, r_2, r_3, \cdot\cdot\cdot, \} is a sequence of rational numbers converging to x, then we define a^x to be the limit \displaytype\lim_{n\to\infty} a^{r_n}.
That way we can assert that both a^xa^y= a^{x+y} and (a^x)^y= a^{xy} are true by definition.
For a completely different point of view, and a completely different proof, we can define e^x
to be the inverse function to ln(x) while defining ln(x) itself by
ln(x)= \int_1^x \frac{1}{t}dt
Let x be a postive real number, y any real number. Then, by that definition,
ln(x^y)= \int_1^{x^y} \frac{1}{t}dt
If y\ne 0, let u= t^{1/y}. Then t= u^y so that dt= y u^{y-1}dy[/tex]. When t= 1, u= 1 and when t= x^y, u= x. So with that change of variable, the integral becomes<br />
ln(x^y)= \int_1^x \frac{1}{u^y}(yu^{y-1}dy)= y\int_1^x \frac{1}{u} du= yln(x).<br />
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If y= 0, then x^y= x^0= 1 so that<br />
ln(x^y)= ln(1)= \int_1^1\frac{1}{t}dt= 0<br />
so we still have ln(x^y)= ln(1)= 0= 0(ln(1))= yln(x).<br />
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Now, if we let w= (e^x)^y, with y\ne 0, we have w^{1/y}= e^x and, since e^x is defined as the inverse function to ln(x), we have x= ln(w^{1/y})= (1/y)ln(w) and so xy= ln(w). Now return to the exponential form: <br />
w= e^{xy} proving that w= (e^x)^y= e^{xy}.<br />
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Of course, if y= 0, then (e^x)^y= (e^x)^0= 1= e^0= e^{x(0)}= e^{xy}
