Assume we have a function $f(x)$, such that $f(a+b)=f(a)f(b)$.

Then, based only on this assumption, $f(3) = f(1)f(1)f(1)$ because $f(1)f(1)f(1) = f(1 + 1 + 1) = f(3)$. For this reason, $f$ must be the exponential function $f(3) = f(1)^3$.

We can also prove that $f(\frac{1}{2}) = \sqrt{f(1)}$, because $f(\frac{1}{2})f(\frac{1}{2}) = f(\frac{1}{2} + \frac{1}{2}) = f(1)$. This forces $2^{\frac{1}{2}}$ to equal $\sqrt{2}$, as it is the only way for $2^{\frac{1}{2}}2^{\frac{1}{2}} = 2^{(\frac{1}{2} + \frac{1}{2})}$.

$f(0) = 1$, as $f(n) = f(n + 0) = f(n)f(0)$ is only possible if $f(0) = 1$.

And then $f(-1) = 1/f(1)$, because $f(1)f(-1) = f(1 - 1) = f(0) = 1$, so $f(-1)$ must result in whatever scalar $c$ such that $cf(1) = 1$, which is $1/f(1)$.

The rules of the notation $a^b$ are designed to achieve the assumption that $a^ba^c = a^{b+c}$, including the mechanic that $a^n$ equals $a$ times itself $n$ times.

Math which exhibits internal consistency is the most useful, as it allows us to prove theorems without worry that it should break down in unverified cases. For this reason, exponentiation is best described by the series: