$$e^x$$ 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 unlimited^{1} 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.
$$ (1+ \frac{x}{N})^N = \sum_{n=0}^{N}{\binom{N}{n} (\frac{x}{N})^n \cdot 1^{Nn}}$$
$$1^{Nn}$$ 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.
$$\sum_{n=0}^{N}{\frac{N!}{n! (Nn)!} \frac{x^n}{N^n} } $$
Cancel $$(Nn)!$$ from $$N!$$ to get
$$\sum_{n=0}^{N}{\frac{N\cdot N1 \cdot \dots Nn+1}{n!} \frac{x^n}{N^n} } $$
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 $$\frac{N}{N}, \frac{N1}{N} \dots \frac{Nn+1}{N}$$. 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 place^{2}.

AKA hyperfinite ↩

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