Equivalence of Definitions of Euler's Number

Theorem
Let $e$ be Euler's Number. The following definitions of $e$ are equivalent.

Proof 1
See Exponential as Limit of Sequence for how $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$ 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 $\displaystyle e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ follows from $\displaystyle \lim_{n \mathop \to \infty} \left({1 + \frac 1 n}\right)^n = e$.

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

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

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:
 * $D_x \left({f \left({x}\right)}\right) = f \left({x}\right)$

From Derivative of Exponential Function:
 * $f \left({x}\right) = e^x$

From Derivative of Inverse Function:
 * $D_x \left({f^{-1} \left({x}\right)}\right) = \dfrac 1 {f^{-1} \left({x}\right)}$

Hence from Derivative of Natural Logarithm Function:
 * $f^{-1} \left({x}\right) = \ln x$

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.

1 implies 2
See Euler's Number: Limit of Sequence Equals Limit of Series.

2 implies 3
See Euler's Number: Limit of Sequence implies Base of Logarithm.

3 implies 4
Let $e$ be the unique solution to the equation $\ln \left({ x }\right) = 1$.

We want to show that $\exp \left({ 1 }\right) = e$, where $\exp$ is the exponential function.

where the final equation holds by hypothesis.

Hence the result.

4 implies 1
Let $e = \exp 1$, where $\exp$ denotes the exponential function.

We want to show that:
 * $\displaystyle \sum_{n = 0}^{\infty} \frac{1}{n!} = e$

Now, by definition of $\exp$:
 * $\displaystyle \sum_{n = 0}^{\infty} \frac{1}{n!} = \exp 1$

And $\exp1 = e$ by hypothesis.

Hence the result.

Also see

 * Equivalence of Definitions of Exponential Function