Equivalence of Definitions of Complex Exponential Function/Power Series Expansion equivalent to Differential Equation

Power Series Expansion implies Solution of Differential Equation
Let $\exp z$ be the complex function defined as the power series:


 * $\exp z := \ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$

Let $y = \ds \sum_{n \mathop = 0}^\infty \frac {z^n} {n!}$.

Then:

We show that $\ds \sum_{n \mathop = 0}^\infty \dfrac {z^n} {n!}$ satisfies the initial condition:


 * $\exp \paren 0 = 1$.

Setting $z = 0$ we find:

That is:

$\exp z$ is the particular solution of the differential equation:
 * $\dfrac {\d y} {\d z} = y$

satisfying the initial condition $\map y 0 = 1$.

Solution of Differential Equation implies Power Series Expansion
Let $\exp z$ be the complex function defined as the particular solution of the differential equation:
 * $\dfrac {\d y} {\d z} = y$

satisfying the initial condition $\map y 0 = 1$.

Let $f: \C \to \C$ be a solution to the differential equation $\dfrac {\d f} {\d z} = f$ with $f \paren 0 = 1$.

Then Holomorphic Function is Analytic shows that $f$ can be expressed as a power series:
 * $\ds \sum_{n \mathop = 0}^\infty a_n \paren {z - \xi}^n$

about any $\xi \in \C$.

When $\xi = 0$, we have for all $n \in \N_{\ge 1}$:

As $a_0 = \dfrac {\map {f^{\paren 0} } 0} {0!} = 1$ by the initial condition, it follows inductively that:


 * $a_n = \dfrac 1 {n!}$

Hence:
 * $\ds \map f z = \sum_{n \mathop = 0}^\infty \dfrac 1 {n!} z^n$