$e^{xy}=(e^{x})^y$ Prove using series. As in the title, I am at loss how to prove it. What I know about exponential function: I know its series, its derivative, it is always positive and that $e^x*e^y=e^{x+y}$. But i have no clue how to show $e^{xy}=(e^{x})^y$. I tried using the trick $\frac{(e^x)^y}{e^{xy}}$ and show it is constant using deriatives but that requires me to show what the derivative of $(e^x)^y$ is, in respect to either x and y. But i cannot do that easily since the sole purpose of this venture is to define $b^x$ for any $b$ and any $x$.
Thanks!
 A: Defining $\exp(x)$ through the series
$$
e^{\,x} \;\mathop  = \limits^{def} \;\sum\limits_{0\, \le \,n} {{{x^{\,n} } \over {n!}}} 
$$
then for the product we get
$$
\eqalign{
  & e^{\,x} e^{\,y} \; = \;\sum\limits_{0\, \le \,n} {{{x^{\,n} } \over {n!}}\sum\limits_{0\, \le \,m} {{{y^{\,m} } \over {m!}}} }  = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,m} {{{x^{\,n} } \over {n!}}{{y^{\,m} } \over {m!}}} }  = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,m} {{1 \over {\left( {n + m} \right)!}}\left( \matrix{
  n + m \cr 
  m \cr}  \right)x^{\,n} y^{\,m} } }  =   \cr 
  &  = \sum\limits_{0\, \le \,s} {\sum\limits_{0\, \le \,m} {{1 \over {s!}}\left( \matrix{
  s \cr 
  m \cr}  \right)x^{\,s - m} y^{\,m} } }  = \sum\limits_{0\, \le \,s} {{1 \over {s!}}\left( {x + y} \right)^{\,s} }  = e^{\,x + y}  \cr} 
$$
hence the inverse, for $x$ real
$$
e^{\, - \,x} e^{\,x}  = e^{\,0}  = 1\quad \left( {e^{\,x} } \right)^{\, - \,1}  = e^{\, - \,x} \quad \left| {\;x \in R} \right.
$$
hence the elevation to an integer power
$$
\left( {e^{\,x} } \right)^n  = e^{\,n\,x} \quad \left| {\,n \in Z} \right.
$$
and the integer root
$$
\left( {e^{\,{x \over n}} } \right)^n  = e^{\,\,x} \quad \left( {e^{\,x} } \right)^{1/n}  = e^{\,\,x/n} \quad \left| {\,0 \ne n \in Z} \right.
$$
and consequently, the elevation to a rational exponent
$$
\left( {e^{\,x} } \right)^{\,{n \over m}}  = e^{\,\,{n \over m}\,x} \quad \left| \matrix{
  \,n,m \in Z \hfill \cr 
  \;0 \ne m \hfill \cr}  \right.
$$
Also, since for $0 \le x$ the sum is definitely positive, the multiplicative inversion tells us that
$$
e^{\, - \,\left| x \right|}  = {1 \over {e^{\,\,\left| x \right|} }}
$$
that is that $\exp(x)$ is definitely positive.
Then,  since the derivative is function itself, that it is increasing all over the domain.
Therefore
$$
{n \over m} < r < \,{{n'} \over {m'}}\,\quad  \Rightarrow \quad \left( {e^{\,x} \,} \right)^{\,\,{n \over m}}  < \left( {e^{\,x} \,} \right)^{\,r}  < \left( {e^{\,x} \,} \right)^{\,\,{{n'} \over {m'}}} 
$$
and from here I think you can conclude by yourself.
