Arithmetic exponentiation
In my perspective, the usual definition of exponentiation is kinda wonky. Exponentiation on $\mathbb{R}$ isn't closed, and exponentiation on $\mathbb{C}$ is multivalued.
I sought for an idea to resolve this. An idea was: Unlike addition or multiplication, exponentiation isn't commutative, so perhaps it isn't even obliged to take arguments from the same object.
Then I reviewed the laws of exponents: $$ 1^y = 1 \\ x^0 = 1 \\ x^1 = x \\ x^{y_1+y_2} = x^{y_1} \cdot x^{y_2} \\ x^{-y} = (x^y)^{-1} \\ (x_1 x_2)^y = (x_1)^y \cdot (x_2)^y \\ x^{y_1 y_2} = (x^{y_1})^{y_2} $$
These rules suggests that, when the exponent object is a ring with unity, the base object is only obliged to be a (multiplicative) group. So instead of having $\text{pow}: R \times R \to R$, I can have $\text{pow}: G \times R \to G$.
So how can I construct $G$ from $R$? It's fairly simple. Given $(R,+,\times)$, forget multiplication to get $(R,+)$, and consider this (abelian) group to be isomorphic to $(G,\times)$. Though usually, $(G,\times)$ wouldn't imbed to $(R,\times)$. Take $e^0 = e^{2 \pi i}$ as a counterexample.
Exponential object
By having analogy to above, I can also conclude that, exponential objects should take different categories as well. Instead of being a bifunctor $\mathsf{pow}: \mathbf{C} \times \mathbf{C} \to \mathbf{C}$, it should be $\mathsf{pow}: \mathbf{D} \times \mathbf{C} \to \mathbf{D}$. A well-known example is $\mathbf{C} = \mathbf{LCHaus}$ and $\mathbf{D} = \mathbf{Top}$.
So overall, I seem to have given a better definition of exponentiation. But are there any downsides?