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.