I am trying to understand how $e$ was originally derived as a solution to the below differential:
$$\dfrac{\partial y}{\partial x} = y$$
I understand how we reach:
$$y=1+x+\dfrac{1}{2}x^2+\dfrac{1}{6}x^3+\cdots=\sum^\infty_{n=0}\dfrac{x^n}{n!}$$
For my sake let's call the above solution to the differential $f(x)$ instead of $y$. It logically follows that $f(1) = e$ where $e$ is some number at which the sum converges for $x=1$. My question is why the functional form is exponential - why do we raise $e$ to $x$ as in the following?
$$f(x) = f(1)^x=e^x$$
I found this Quora post but unfortunately got lost where the summation notation was rearranged as I'm unfamiliar with summation laws. I see we that somehow we separate out the binomial theorem from the numerators but cannot understand how this happens.
Any help understanding this would be greatly appreciated!
Use the differential equation $f'(x) = f(x), f(0)=1$ to show $f(x+y) = f(x) f(y)$. Then $f(x) = f(1)^x$ for $x$ integer. Then $f(x) = f(1)^x$ for $x$ rational. And finally define irrational exponent by continuity to get $f(x) = f(1)^x$ for all $x$