Proof that $\sum_{n=0}^{\infty}\frac{x^n}{n!}=\lim_{n\to \infty}\left( 1+\frac{x}{n}\right)^n$ Assume that someone in another planet has discovered that the function $f(x)=\frac{1}{x}$ has a bijective primitive denoted by $F(x)$. How would he prove that the inverse of this primitive $E(x)=F^{-1}(x)$ satisfies:
1) $E(x)=\lim_{n\to \infty}\left( 1+\frac{x}{n}\right)^n$
2)
$E(x)=\sum_{n=0}^{\infty}\frac{x^n}{n!}.$
3) $E(x+y)=E(x)E(y)$
 A: You have $$F(x)=\int_1^x\frac{1}{z}dz$$ then you show 
$$F(xy)=F(x)+F(y)$$ by 
$$\int_1^{xy}\frac{1}{z}dz=\int_1^x\frac{1}{z}dz+\int_x^{xy}\frac{1}{z}dz$$
$$=\int_1^x\frac{1}{z}dz+\int_1^y\frac{1}{z}dz$$ where a change of variable has been used in the last line. This is equivalent to 3). Next you can show $$E^{\prime}(x)=E(x)$$ from $F^{\prime}=\frac{1}{x}$ and $F(E(x))=x$, and the chain rule. To derive 2), use the theory of McLauren series. The first two are known equivalence, but I dont know how to motivate 1) from the current standpoint.
A: For 1), prove the property in Rene Schipperus's answer first, that $F(xy) = F(x) + F(y)$, which also implies $F(x^n) = nF(x)$. 
Now note that $F(\lim_{n \to \infty} \left( 1 + \frac{x}{n})^n\right)$ is 
$$
\lim_{n \to \infty} F\left( \left(1 + \frac{x}{n}\right)^n\right)
$$
because $F$ is continuous, being the primitive of a continuous function. Now by property 3), this is
$$
\lim_{n \to \infty} n F\left(1 + \frac{x}{n}\right) = \lim_{n \to \infty} \frac{F(x+n) - F(n)}{1/n}
$$
which by L'Hôpital is
$$
\lim_{n \to \infty} \frac{\frac{1}{x+n}-\frac{1}{n}}{\frac{-1}{n^2}} = \lim_{n \to \infty} \frac{-n^2}{x+n} +n = \lim_{n \to \infty} \frac{nx + n^2 - n^2}{x + n} = \lim_{n \to \infty} \frac{nx}{n(1 + x/n)} = x.
$$
So the inverse function to $F$ is given by the limit formula.
A: Suppose this person from another planet knows about a unique complete ordered archimedian field, and from such set she is able to develop the modern theory of single-variable Calculus, including integral of continuous functions and derivatives, L’Hôpital, power series, and uniqueness of antiderivatives modulo additive constants.
Then (3) is shown from the definition as explained in another answer, (1) follows as usual by tanking log and applying L’Hopital, and (2) requires more care.
First show that the power series $S(x)$ has infinite radius of convergence, and use the fact that inside the radius of convergence a power series converges so fast that it can be integrated term by term, so $S(x)$ is an antiderivative of itself, and conclude that $S’=S$.
Now using the chain rule and $S’=S$ you conclude that the derivative of $F\circ S$ is $1$, from which you conclude that $F(S(x))=x$, so $S=E$.
