Proof that $e^{ab} = \left(e^a\right)^b$, using the series for $e^x$ I want to prove that 
$$e^{ab}=\left(e^{a}\right)^{b}$$where 
$$a,b \in \mathbb{R}$$ using the infinite series for $$e^{x}$$
My attempt:
$$e^{ab}=\sum_{n=0}^{\infty}\frac{(ab)^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(a)^{n}(b)^{n}}{n!}=\sum_{n=0}^{\infty}\frac{(a)^{n}}{n!}b^{n}=1+(a)b+\left(\frac{a^{2}}{2}\right)b^{2}+\left(\frac{a^{3}}{6}\right)b^{3}+\cdots$$
At this point I don't know how to proceed. I don't see how I could get b out of the series and have it as a power that the series is raised to. I thought that it resembled the geometric series, but that did not seem to lead anywhere.
 A: The problem with trying to prove this is that you need a rigorous definition for $x^y$ where $x$ can be any positive number and $y$ can be any real number. Now, we already have a rigorous definition for $\exp(y)=e^y$ for any $y\in\Bbb{R}$ as $\sum_{n=0}^\infty \frac{y^n}{n!}$. Therefore, I usually people write $x^y$ in terms of the exponential function:
$$x^y=\exp(y\ln x)$$
Thus, the easiest way to prove $(e^a)^b$ is to just use this definition of $x^y$:
$$(e^a)^b=\exp(b\ln e^a)=\exp(ba)=e^{ab}$$
I know that's not really the series solution you wanted, but I still think it is helpful to see this kind of proof.
A: If we assume the "meaning" of $e^a$ and if $b\in \mathbb{Q}$, $b=\frac{p}{q}$, then we know what does it mean to $\textit{raise a number by the power $p$ and taking the $q-th$ root of that.}$ That is to say, $(e^a)^p=e^a...e^a=e^{ap}$, and $(e^{ap})^{\frac{1}{q}}$ is a number which gives $e^{ap}$ when multiplied $q$-times, and we are done.
Now if the number $b$ is not a rational, then consider the following number (can be defined in this way, uses l.u.b. property of $\mathbb{R}$) $$(e^a)^b=\sup\{(e^a)^x\,\,|\,\, x\in \mathbb{Q}\text{ and }x<b\}=\sup\{e^{ax}\,\,|\,\, x\in \mathbb{Q}\text{ and }x<b\}=e^{ab}.$$
