Exponential of Real Number is Strictly Positive/Proof 5/Lemma

Theorem
Let $x$ be a real number.

Let $\exp$ denote the (real) Exponential Function.

Then:
 * $\forall x \in \R: \exp x \ne 0$

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

That is, suppose $\exp$ satisfies:
 * $ (1): \quad D_x \exp x = \exp x$
 * $ (2): \quad \exp \left({0}\right) = 1$

on $\R$.

that $\exists \alpha \in \R: \exp \alpha = 0$.

Suppose that $\alpha > 0$.

Let $J = \left[{0 \,.\,.\, \alpha}\right]$.

From Exponential Function is Continuous, $\exp$ is continuous on $J$.

From Max and Min of Function on Closed Real Interval:
 * $\exists K \in \R: \forall x \in J: \left\vert{\exp x}\right\vert < K$

Then, $\forall n \in \N : \exists c_n \in J$ such that:

So $\forall n \in \N :1 \le K \dfrac{\alpha^{n}}{n!}$

That is, dividing both sides by $K$:
 * $\forall n \in \N: \dfrac 1 K \le \dfrac{\alpha^{n}}{n!}$

But from Power over Factorial, $\dfrac{\alpha^{n}}{n!} \to 0$.

This contradicts our assumption.

The same argument, mutatis mutandis proves the result for $\alpha < 0$.

By hypothesis $(2)$:

Hence the result.