Euler's formula for e^x

We want to prove the identity $$e^x=\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{x}{n})}^n.$$ If we consider the Taylor series for $e^x$, on the left hand side we have $$e^x=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\cdots$$ On the right, we can expand out the product using the Binomial Theorem $${(1+\frac{x}{n})}^n=\sum _{i=0}^n\frac{n(n-1)\cdots (n-i+1)}{i!}{\left(\frac{x}{n}\right)}^i=\sum _{i=0}^n\frac{x^i}{i!}\frac{n(n-1)\cdots (n-i+1)}{n^i}.$$ The coefficient of $x^i$ equals $$\frac{1}{i!}\frac{n(n-1)\cdots (n-i+1)}{n^i}=\frac{1}{i!}\frac{n}{n}\frac{n-1}{n}\cdots \frac{n-i+1}{n}.$$ As $n\to \mathrm{\infty }$ each term here containing $n$ converges to $1$ giving $$\underset{n\to \mathrm{\infty }}{lim}{(1+\frac{x}{n})}^n=\sum _{i=0}^{\mathrm{\infty }}\frac{x^i}{i!},$$ or exactly the Taylor series for $e^x$.

You might think this argument has made somewhat cavalier use of limits. The following argument, adapted from [1, pg 272], is close to Euler's original. He was even more bold in his use of limits!

Thinking about $e^x$ near $x=0$, and its gradient, we see that, for infinitely small $ϵ$ $$e^{ϵ}=1+ϵ$$ This is illustrated above. Let $x$ be any given number, then $N=x/ϵ$ is infinitely large. So %$$\text{Missing end{eqnarray*}}$$ $$e^x=e^{Nϵ}=(e^{ϵ}{)}^N=(1+ϵ{)}^N={(1+\frac{x}{N})}^N.$$ Written in terms of limits, the infinitely large $N$ would give us (1).

Reference

[1] C.H. Edwards, The Historical Development of the Calculus, Springer-Verlag, 1979.