How do we show that the function which is its own derivative is exponential? In my calculus class, to show that $\frac{d}{dx}e^x=e^x$ we did something like this:

$$\lim_{h \to 0} \frac{a^{x+h} - a^x}{h} = a^x \lim_{h \to 0} \frac{a^h-1} h,$$

and then we defined $e$ to be the base $a$ that makes the limit

$$\lim_{h \to 0} \frac{a^h-1}{h}$$

equal to $1$. Now, my question is this: how can we know beforehand that the function which is its own derivative is an exponential function? In other words, if we have 

$$ \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} = f(x),$$

how do we know $f(x)$ is of the form $ka^x$ (ignoring the trivial case $f(x) = 0$)? How can we show that there aren't functions of other forms that give the same result?
Hint. If $f$ is such a function then calculating the derivative of $f(x)/e^x$ provides useful information about $f$.
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We want an expression which when differentiated yields itself.
The first possible case is when $f(x)=0.$

Second, let's solve the initial value problem $f(0)=1$ and $f(x)=f'(x)$.
Now taking $f(x)=1,$ and then integrating the previous function and adding the result to obtain a better solution to the initial value problem, we get,
$f(x)=1 + x$, and then further integrating the second element, and repeating we get,
$f(x)=1 + x + (x^2)/2.$
Repeating the process will eventually yield the taylor series expansion for the exponential function as $f(x)=\sum_{i=0}^n (x^n)/k!$
And then we define the nth root of this expression as being the constant e.
Also, by differentiating the inverse of the function $f(x)$, we arrive at $(f^-1)'(x)=1/x$, which is the natural logarithm.

The essence of the exponential function with base e is that it is its own derivative and as such can be taken as its definition.

PS: We can also define e to be the positive upper limit of integration in the integral definition of the natural logarithm, and deduce that $e^x$ is its own derivative.