In the 18th century, Leonard Euler found this formula to write explicitly exponential :

$$x \in \mathbb{C} \qquad e^x = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$

But how to find it ? It is what I'm going to show you.

Proof

This proof is the one that I found on my own (but isn't most probably the one that Euler found) and only uses exponential properties and the assumption that \(e^x\) can be written like a polynomial of infinite degree, i.e. these 3 equalities :

$$\begin{gather*}
\exp' = exp \\
e^0 = 1 \\
e^x = \sum_{k = 0}^{\infty} a_k x^k
\end{gather*}$$

From these equalities, let's use a proof by induction with the following statement :

$$\forall n \in \mathbb{N} \quad H_n : a_n = \frac{1}{n!}$$

that leads us to the following reasoning :

Bases :
For \(n = 0\), \(e^0 = a^0 = 1 = \frac{1}{0!}\). So, \(H_0\) holds.

Inductive step :
As \(\exp = \exp'\), we can write the following :$$\sum_{k = 0}^{\infty} a_k x^k = (\sum_{k = 0}^{\infty} a_k x^k)' = \sum_{k = 0}^{\infty} (k+1) a_{k+1} x^k$$which means that for \(m \in \mathbb{N}\),

$$a_m = (m+1)a_{m+1} \Leftrightarrow a_{m+1} = \frac{a_m}{m+1}$$

And, as we assume \(H_m\) holds,

$$a_{m+1} = \frac{a_m}{m+1} = \frac{\frac{1}{m!}}{m+1} = \frac{1}{(m+1)!}$$

Consequently, \(H_m \Rightarrow H_{m+1}\).

Conclusion :
\(H_n\) holds \(\forall n \in \mathbb{N}\). Thus :

$$e^x = \sum_{k = 0}^{\infty} \frac{1}{k!} x^k = \sum_{k = 0}^{\infty} \frac{x^k}{k!}$$