Why does the power series expressing e^x have the form of a constant raised to x ? This question is probably very basic, but I've been away from school for a while and the answer eludes me.
I was tempted to prove that d/dx(e^x) = (e^x) for old times sake and that was easy enough. I just expressed e^x as a power series where n goes from 0 to infinity for ((x^n)/n!).
During the derivation I started to wonder, how did they know that a power series where n goes from 0 to infinity for A(subscript n)*X^n would converge to the form of (someconstant)^x.
Is there some theorem that proves this?
This question is pretty probably trivial for the hardcore math types but it's been bothering me, so I thought I'd ask :-) ....
 A: Some people would say that your question is trivial because we define the power function by a^x = exp(x log a).
However, that's not a very satisfying answer.
Clearly one wants the power series for exp(x) to satisfy exp(z+w) = exp(z) exp(w), and exp(0) = 1, from what we know the power function should be if z and w are integers.    (I'm writing exp(x), not e^x, because I'm assuming exp(x) hasn't yet been shown to have this property.)
So say exp(x) = a0 + a1 x + a2 x^2 + a3 x^3 ... is a formal power series satisfying exp(z+w) = exp(z) exp(w).
Then since exp(0) = 1, we must have a0 = 0.
So exp(x) = 1 + a1 x + a2 x^2 + a3 x^3 + ...; therefore
exp(2x) = exp(x) exp(x) = (1 + a1 x + a2 x^2 + a3 x^3 + ...) (1 + a1 x + a2 x^2 + a3 x^3 + ...)
and expanding the rightmost member of this equation as a formal power series,
exp(2x) = 1 + 2a1 x + (2a2 + a1^2) + (2a3 + 2 a2 a1) x^3 + ...
However, exp(2x) = 1 + 2a1 x + 4a2 x^3 + 8a3 x^3 + ... by substituting 2x into the formal power series for exp(x).
By equating the coefficients of x, x^2, and x^3, you get
2 a1 = 2 a1
2 a2 + a1^2 = 4 a2
2 a3 + 2 a2 a1 = 8 a3
and so on.  The first equation tells you nothing.  The second tells you a1^2 = 2a2, so a2 = a1^2/2.  The third becomes
2 a3 + 2 (a1^2 / 2) a1 = 8 a3
from which you get a1^3 = 6 a3, and a3 = a1^3/6.  The pattern here continues, with an = a1^n/n!, as can be proven by induction.
This gives the series
exp(x) = 1 + a1 x + a1^2/2! x^2 + a1^3/3! x^3 + ...
and now we just have to choose a1.  We pick 1 just because it's simple to do so.
A: One answer is that the power series sum x^n/n! is used to define exponentials, not the other way around.  The way to understand this is in terms of the defining property of the exponential, which is that e^{x+y} = e^x e^y.  It turns out that this property is more or less equivalent to the property that d/dx e^x = e^x, and this property in turn is equivalent to having a certain Taylor series expansion.
Edit:  Let me sketch the arguments here.  In one direction, if e^{x+y} = e^x e^y then d/dx e^x = lim (e^{x+h} - e^x)/h = e^x lim (e^h - 1)/h.  Since lim (e^h - 1)/h is just the derivative of e^x evaluated at 1, we can pretend that this exists and normalize it to 1, and then d/dx e^x = e^x as desired. 
In the other direction, if d/dx e^x = e^x then d/dx e^{x+y} = e^{x+y} for fixed y.  On the other hand, the space of functions satisfying d/dx f(x) = f(x) has to be one-dimensional (just take this on faith; it's more trouble to justify than it's worth right now), so e^{x+y} has to be a constant multiple of e^x, say e^{x+y} = Ke^x, and setting x=0 gives K = e^y as expected. 
As for how these both relate to Taylor series, if d/dx e^x = e^x this implies that every Taylor coefficient of e^x is equal.  
Edit #2:  Anyway, one lesson to take away from all this is that there are multiple equivalent definitions of e^x, and the way you go about proving any of its properties depends strongly on what definition you use.  There are good technical reasons to start with the power series definition: 1) you are automatically guaranteed that the function is well-defined everywhere, and 2) the power series definition makes sense in a very general context, such as over the complex numbers or with matrix inputs.
