Equivalence of Definitions of Euler's Number/Proof 1

Proof
See Equivalence of Definitions of Real Exponential Function: Inverse of Natural Logarithm implies Limit of Sequence for how $\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = e$ 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 $\ds e = \sum_{n \mathop = 0}^\infty \frac 1 {n!}$ follows from $\ds \lim_{n \mathop \to \infty} \paren {1 + \frac 1 n}^n = e$.

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

Let us consider the series $\ds \map f x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$.

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

We differentiate $\map f x$ $x$ term by term (justified by Power Series is Differentiable on Interval of Convergence), and get:

Thus we have:
 * $\map {D_x} {\map f x} = \map f x$

From Derivative of Exponential Function:
 * $\map f x = e^x$

From Derivative of Inverse Function:
 * $\map {D_x} {\map {f^{-1} } x} = \dfrac 1 {\map {f^{-1} } x}$

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

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.