Exponential Function is Well-Defined/Real/Proof 5

Theorem
Let $x \in \R$ be a real number.

Let $\exp x$ be the exponential of $x$.

Then $\exp x$ is well-defined.

Proof
This proof assumes the definition of $\exp$ as the solution to an initial value problem.

That is, suppose $\exp$ solves:
 * $ (1): \quad \dfrac {\mathrm d} {\mathrm d x} y = f \left({x, y}\right)$
 * $ (2): \quad \exp \left({ 0 }\right) = 1$

on $\R$, where $f \left({x, y}\right) = y$.

From Derivative of Exponential Function: Proof 4, the function $\phi : \R \to \R$ defined as:
 * $\displaystyle \phi \left({ x }\right) = \sum_{k \mathop = 0}^{\infty} \frac{x^k}{k!}$

satisfies $\phi ' \left({ x }\right) = \phi \left({ x }\right)$.

So $\phi$ satisfies $(1)$.

From Exponential of Zero: Proof 3:
 * $\phi \left({ 0 }\right) = 1$

So $\phi$ satisfies $(2)$.

Thus, $\phi$ is a solution to the initial value problem given.

From Exponential Function is Continuous: Proof 5  and $(1)$, $\phi$ is  continuously differentiable on $\R$.

Since $f \left({x, \phi}\right) = \phi$, $f$ is continuously differentiable on $\R^{2}$.

Thus, from Uniqueness of Continuously Differentiable Solution to Initial Value Problem, this solution is unique.