Power Series Expansion for Exponential Function

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

Then:

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

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

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

From Taylor's Theorem, we know that
 * $\ds \exp x = 1 + \frac x {1!} + \frac {x^2} {2!} + \cdots + \frac {x^{n - 1} } {\paren {n - 1}!} + \frac {x^n} {n!} \map \exp \eta$

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

Hence:

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

The result follows.