Braindead Way to Derive Taylor Series of Exponential Function

$$e^x$$ 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.