Proof of exponential theorem I am currently reading Joseph Edwards' differential​ calculus for beginners. I am a beginner, in am in 10th. 
$$a^x = 1 + x\ln a + \frac{x^2}{2!} (\ln a)^2 + \frac{x^3}{3!} (\ln a)^3...$$ I know it makes sense because if you plug in $e=a$ you get the maclaurin expansion for $e^x$. But what about a rigorous proof. I have tried messing around with it. I can't see a connection for natural logs be appearing in an exponent like $a^x$ without $e$ being in it? Please help. Thanks in advance.
 A: Note that 
$$
a^x=e^{x\ln a}=\sum_{k=0}^\infty\frac{x^k(\ln a)^k}{k!}
$$
where the second equality comes from knowledge of the Maclaurin series for $e^x$
A: It comes from the Taylor's expansion for $a^x$ since $(a^x)'=a^x\log a$, $(a^x)''=a^x\log^2 a$ and so on.
Note
$$a^x = e^{x\ln{a}}\implies (a^x)'=(e^{x\ln{a}})'=e^{x\ln{a}}\ln a=a^x \ln a$$
A: Let $a^x = y$.
$x \ln a = \ln y$
$e^{x\ln a}= y$  
$\therefore a^x = e^{x\ln a}$
$a^x = {(e^x) }^{\ln a}$
$a^{\frac {x}{\ln a}} = e^x $
Substituting $a^{\frac {x}{\ln a}} = e^x$ into the Maclaurin expansion for $e^x$ should give you the same expansion as your original one.
(Note: this may be wrong so please pardon me. I am also in the $10^{th}$ grade and I am supposed to learn this after $2$ years).
A: By Taylor series we have: 
$$f(x)=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$
with $f(x)= a^x =e^{x\ln a}$ see that 
$$f'(x) =\ln a ~e^{x\ln a}= \ln a\cdot f(x),~~f^{(2)}(x)= (\ln a)^{2} f(x)\cdots\cdots , f^{(n)}(x)= (\ln a)^{n} f(x)$$
that is $$f(0) =1,~~f'(0) =\ln a ,~~f^{(2)}(0)= (\ln a)^{2} \cdots\cdots , f^{(n)}(0)= (\ln a)^{n} $$
