\(e^x\) 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. ↩