Our goal in this post is to extract the one true meaning of the exponential function: that it converts addition to multiplication. We will begin to make our case with the following observations.
When we first learn about expressions like $a^3$, we think about them as “repeated multiplication.” More fundamental is the idea that $a^{x+y} = a^x\cdot a^y$, which implies the first fact. More generally, we have a certain number of rules for what $a^q$ should be whenever $q$ is a rational number. These all follow from our fundamental idea. For example, $a^1 = a^{\frac{1}{2} + \frac{1}{2}} = (a^{\frac{1}{2}})^2$, which tells us all about the square root rule.
Notice that we have used absolutely nothing about $a^x$ other than a mild generalization of the fact that it is "repeated multiplication" for natural numbers. There is now a mathematical swindle that the crafty mathematicians of history have employed. We will join their ranks: assume that there exists some function $f$ is some function where $f(x+y) = f(x)\cdot f(y)$.
From now on, calculus will tell us everything we need. Let's assume that $f$ can be differentiated infinitely many times (actually we only need continuous but we won't pursue that here--the solution is no fun anyway.) Now, differentiating with respect to $t$, we get the following equation:
$$\frac{d}{dt}f(t+x) = \frac{d}{dt}f(t)f(x) = f(x)\frac{d}{dt}(f(t)).$$
Setting $t = 0$, we see that $f^{\prime}(x) = f^{\prime}(0)f(x)$. This is the defining property of the exponential map! Let's write $a = f^\prime{0}$. First notice that since $f(x+y) = f(x)f(y)$, it's automatic that $f(0) = 1$. Then, the Maclaurin expansion of $f$ precisely agrees with that of $C e^{ax}$ (I'll leave this as a little exercise).
For those that know ordinary differential equations, we know that the unique solution for the above ODE is already of the form $e^{ax}$. This allows us with getting away without the assumption that the exponential has all derivatives.
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