A sublime proof of Euler’s formula

A proof of Euler’s formula using the Leibniz product rule.

The most remarkable formula in mathematics.-Feynman

The complex exponential may be defined as follows:

\[\begin{equation} \text{exp}: \mathbb{C} \rightarrow \mathbb{C} \\ z \mapsto \sum_{n \geq 0} \frac{z^n}{n!} \tag{1} \end{equation}\]

Using the Cauchy Product and Mertens’ theorem, this implies that:

\[\begin{equation} \forall a, b \in \mathbb{C}, \text{exp}(a) \cdot \text{exp}(b) = \text{exp}(a + b) \tag{2} \end{equation}\]

Now, let’s consider the function:

\[\begin{equation} \forall t \in \mathbb{R}, f(t) = e^{-it} \cdot (\cos t +i\sin t ) \tag{3} \end{equation}\]

By the Leibniz product rule:

\[\begin{equation} \forall t \in \mathbb{R}, f'(t) = -i \cdot e^{-it} \cdot (\cos t +i\sin t ) + e^{-it} \cdot (i \cos t -\sin t ) = 0 \tag{4} \end{equation}\]

Thus, \(f\) is constant everywhere.

Now, given that:

\[\begin{equation} f(0)=1 \implies \forall t \in \mathbb{R}, f(t)=1 \tag{5} \end{equation}\]

we may conclude that:

\[\begin{equation} \forall t \in \mathbb{R}, e^{it} = \cos t + i \sin t \tag{6} \end{equation}\]

QED.