Suppose I knew nothing about the function $e^x$. If I wanted to find a power series that was its own derivative, I would logically start with the constant term, and first start by setting it to $1$. Then, the next term should be the antiderivative of the first term, giving me $x$. Doing this again would give me $\frac{x^2}2$. Repeating this process over and over again, I would get
$$1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\ldots=\sum_{k=0}^{\infty}\frac{x^k}{k!}$$
Graphing a few terms of this, I might notice that this looks more and more like an exponential the more terms I graph. If I prove that this function satisfies the exponential relationship $f(x+y) = f(x)f(y)$, I would be able to prove that this series is an exponential function. How would I prove this? After this, how would I prove that the base of this exponential function is $e$, which is defined as $\displaystyle{\lim_{n \to \infty}} (1+\frac1n)^n$?
Edit: After expanding $(1+x)(1+y)$ and $(1+x+\frac{x^2}2)(1+y+\frac{y^2}2)$, I can see how extra terms get taken care of when the next degree is added. However, my second question still stands.
Best Answer
The first answer lies in Cauchy products and the binomial theorem, which show that
$$e^xe^y=\sum_{n=0}^\infty\sum_{k=0}^\infty\frac{x^ny^k}{n!k!}=\sum_{n=0}^\infty\frac1{n!}\sum_{k=0}^n\binom nkx^{n-k}y^k=\sum_{n=0}^\infty\frac{(x+y)^n}{n!}=e^{x+y}$$
As per the second question, the proof can be lengthy and various, and many are outlined in
The definition of e by limits of $(1+1/n)^n$ through series expansion.