How do I prove $\exp(1)=e$ if $\exp$ is defined in a series expansion?

Let $N\in\Bbb N$. Then, for $n\geqslant N$,\begin{align}\left(1+\frac1n\right)^n&=\sum_{k=0}^n\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)\\&\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right).\end{align}Therefore,$$e\geqslant\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)$$and so$$e\geqslant\lim_{n\to\infty}\sum_{k=0}^N\frac1{k!}\left(1-\frac1n\right)\cdots\left(1-\frac{k-1}n\right)=\sum_{k=0}^N\frac1{k!}.$$Since this takes place for each $N\in\Bbb N$,$$e\geqslant\sum_{k=0}^\infty\frac1{k!}=\exp(1).$$

This appears as Theorem 8.1.6 in Strichartz's The Way of Analysis. The idea is the following:

• First use the series definition to develop some basic properties of $\exp$: it is strictly positive on $\mathbb{R}$, it is its own derivative, it is strictly increasing, it satisfies $(\exp x)^n = \exp(nx)$.

• Use the inverse function theorem to show that $\exp$ has an inverse function, call it $\ln$, which is differentiable. Use the chain rule to show $\frac{d}{dx} \ln x = \frac{1}{x}$. Also verify that $\ln(1)=0$ and $\ln (x^n) = n \ln x$.

• Write $$ \lim_{n \to \infty} \ln \left(1 + \frac{1}{n}\right)^n = \lim_{n \to \infty} \frac{\ln\left(1 + \frac{1}{n}\right) - \ln(1)}{\frac{1}{n}} = \left.\frac{d}{dx}\right|_{x=1} \ln x = 1.$$

• Take $\exp$ of both sides and use the continuity of $\exp$ to conclude.