Power Series Expansion for Exponential Function

Real Numbers
Let $$\exp x$$ be the exponential function.

Then $$\forall x \in \mathbb{R}: \exp x = \sum_{n=0}^\infty \frac {x^n} {n!}$$.

Proof for Real Numbers
Let us take the definition of the exponential function, we have

From Differential of Exponential Function, we have $$f^{\prime} \left({\exp x}\right) = \exp x$$.

Thus it follows that $$\forall n \in \mathbb{N}: f^{\left({n}\right)} \left({\exp x}\right) = \exp x$$.

Since $$\exp 0 = 1$$, the Taylor series expansion for $$\exp x$$ about $$0$$ is given by $$\exp x = \sum_{n=0}^\infty \frac {x^n} {n!}$$.

From Power Series over Factorial, we know that this power series expansion converges for all $$x \in \mathbb{R}$$.

From Taylor's Theorem, we know that $$\exp x = 1 + \frac {x} {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n-1}} {\left({n-1}\right)!} + \frac {x^n} {n!} \exp \left({\eta}\right)$$, where $$0 \le \eta \le x$$.

Hence:

$$ $$ $$

So the partial sums of the power series converge to $$\exp x$$.

The result follows.