Power Series Expansion for Exponential Function

Theorem
Let $\exp x$ be the exponential function.

Then:
 * $\displaystyle \forall x \in \R: \exp x = \sum_{n \mathop = 0}^\infty \frac {x^n} {n!}$

Proof
From Higher Derivatives of Exponential Function, we have:
 * $\forall n \in \N: f^{\left({n}\right)} \left({\exp x}\right) = \exp x$

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

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

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

where $0 \le \eta \le x$.

Hence:

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

The result follows.