Equivalence of Definitions of Euler's Number

Theorem
The two definitions of Euler's number $e$ as given by:


 * $$e = \lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n$$


 * $$e = \sum_{n=0}^\infty \frac 1 {n!}$$

are equivalent to the definition of $$e$$ as the number satisfied by $$\ln e = 1$$.

Thus all definitions for $$e$$ are equivalent.

Proof
See Exponential as the Limit of a Sequence for how $$\lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = e$$ follows from the definition of $$e$$ as the number satisfied by $$\ln e = 1$$.

See Euler's Number: Limit of Sequence implies Limit of Series for how $$e = \sum_{n=0}^\infty \frac 1 {n!}$$ follows from $$\lim_{n \to \infty} \left({1 + \frac 1 n}\right)^n = e$$.

Now suppose $$e$$ is defined as $$e = \sum_{n=0}^\infty \frac 1 {n!}$$.

Let us consider the series $$f \left({x}\right) = \sum_{n=0}^\infty \frac x {n!}$$.

From Series of Power over Factorial Converges, this is convergent for all $$x$$.

We differentiate $$f \left({x}\right)$$ WRT $$x$$ term by term (justified by Power Series Differentiable on Interval of Convergence), and get:

$$ $$ $$ $$

Thus we have $$D_x \left({f \left({x}\right)}\right) = f \left({x}\right)$$ and thus from Differential of Exponential Function it follows that $$f \left({x}\right) = e^x$$.

From Derivative of an Inverse Function we get that $$D_x \left({f^{-1} \left({x}\right)}\right) = \frac 1 {^{-1} \left({x}\right)}$$.

Hence from Basic Properties of Natural Logarithm it follows that $$f^{-1} \left({x}\right) = \ln x$$.

It follows that $$e$$ can be defined as that number such that $$\ln e = 1$$.

Hence all the definitions of $$e$$ as given here are equivalent.