$$e^x$$ was first defined by a formula for continuous compound interest:

$e^x := \lim_{N \to \infty} (1+ \frac{x}{N})^N$

Let’s use this to get that $$e^x = \sum_{n=0}^{N}{\frac{x^n}{n!}}$$

First, remove the limit and replace it with an unlimited1 integer $$N$$.

That gives us the expression $$(1+ \frac{x}{N})^N$$. Which is just a polynomial with an unlimited number of terms. We can expand it by the binomial theorem.

$(1+ \frac{x}{N})^N = \sum_{n=0}^{N}{\binom{N}{n} (\frac{x}{N})^n \cdot 1^{N-n}}$

$$1^{N-n}$$ is just 1, so we can drop it because it’s the identity for multiplication. Which gives us

$\sum_{n=0}^{N}{\binom{N}{n} (\frac{x}{N})^n }$

We can expand $$\binom{N}{n}$$ by using its definition and move the $$n$$ exponent into each term.

$\sum_{n=0}^{N}{\frac{N!}{n! (N-n)!} \frac{x^n}{N^n} }$

Cancel $$(N-n)!$$ from $$N!$$ to get

$\sum_{n=0}^{N}{\frac{N\cdot N-1 \cdot \dots N-n+1}{n!} \frac{x^n}{N^n} }$

Now for the only real bit of cleverness. N is unlimited, so $$\frac{N - k}{N}$$ where k is limited is infinitely close to 1. Infinite closeness is an equivalence relation, which is almost as good as true equality. In this case, we use it to cancel out $$\frac{N}{N}, \frac{N-1}{N} \dots \frac{N-n+1}{N}$$. This leaves behind

$\sum_{n=0}^{N}{\frac{x^n}{n!}} \blacksquare$

I read Euler recently and it turns out this is how he found the Taylor Series in the first place2.

1. AKA hyperfinite

2. Euler didn’t have rigorous rules of unlimited integers, but his intuition was on point.