Comment by tsimionescu

I forgot to respond to this:

> And it seems that way. Say that you ignore taylor series formulation, skip it completely, and define the polar form to be r(10)^ix = cos(x)+isin(x), where r is radius and x is angle. JUST BECAUSE YOU CAN.

The thing is, you can't do this. Your numbers will not work out correctly. For example, (e^(pi*i))^i = 1/e^pi is a direct consequence of how exponentiation works and the definition of i. If you define 10^(xi) = cos x + i sin x, you will get:

  (e^(pi\*i))^i = 
    = (10^(log_10 e \* pi \* i))^i
    = (cos (log_10 e \* pi) + i sin (log_10 e \* pi)) ^ i
    = 10^(log_10 $exp)i
    = cos $exp + i sin $exp

I very much doubt that sin(cos (log_10 e * pi) + i sin (log_10 e * pi)) is 0, so this number can't possibly be equal to the real number 1/e^pi. So, with your definition, the property (a^x)^y = a^(x*y) doesn't hold for all a, x, y. So, your definition doesn't represent the exponential function at all.