${e}^{r}$ is irrational for $r\in \mathbb{Q}\setminus \{0\}$
We here present a proof of the following theorem:
Theorem.
${e}^{r}$ is irrational for all $r\mathrm{\in}\mathrm{Q}\mathrm{\setminus}\mathrm{\{}\mathrm{0}\mathrm{\}}$
To begin with, note that it is sufficient to show that ${e}^{u}$ is irrational for any positive integer (http://planetmath.org/NaturalNumber)^{1}^{1}In this entry, $\mathbb{N}:=\{1,2,3,\mathrm{\dots}\}$ and ${\mathbb{N}}_{0}:=\mathbb{N}\cup \{0\}$. $u$ (for if ${e}^{r}={e}^{\frac{u}{v}}$ were rational, so would ${({e}^{\frac{u}{v}})}^{v}={e}^{u}$). Next, we look at some simple properties of polynomial ${f}_{n}(x):={\displaystyle \frac{{x}^{n}{(1x)}^{n}}{n!}}$:

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${f}_{n}(x)={\displaystyle \frac{1}{n!}}{\displaystyle \sum _{i=n}^{2n}}{c}_{i}{x}^{i}$, with ${c}_{i}\in \mathbb{Z}$ for all $i$.

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${f}_{n}^{(k)}(0)$ and ${f}_{n}^{(k)}(1)$ are integers for all $k\in {\mathbb{N}}_{0}$: as $0$ is a root (http://planetmath.org/Root) of order $n$, ${f}_{n}^{(k)}(0)=0$ unless $n\le k\le 2n$, in which case ${f}_{n}^{(k)}(0)=\frac{k!}{n!}{c}_{k}$, an integer. Since ${f}_{n}^{(k)}(x)={(1)}^{k}{f}_{n}^{(k)}(1x)$, the same is true for ${f}_{n}^{(k)}(1)$.

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For all $$ we have $$.
Now we can readily prove the theorem:
Proof.
Assume that ${e}^{u}=\frac{a}{b}$ for some $(a,b)\in {\mathbb{N}}^{2}$ and let
$${F}_{n}(x):=\sum _{k=0}^{\mathrm{\infty}}{(1)}^{k}{u}^{2nk}{f}_{n}^{(k)}(x),$$ 
which is actually a finite sum since ${f}_{n}^{(k)}(x)=0$ for all $k>2n$. Differentiating ${F}_{n}(x)$ yields ${F}_{n}^{\prime}(x)={u}^{2n+1}{f}_{n}(x)u{F}_{n}(x)$ and thus:
$$\frac{d}{dx}\left[{e}^{ux}{F}_{n}(x)\right]=u{e}^{ux}{F}_{n}(x)+{e}^{ux}{F}_{n}^{\prime}(x)={u}^{2n+1}{e}^{ux}{f}_{n}(x).$$ 
Now consider the sequence^{}
$${({w}_{n})}_{n\in \mathbb{N}}:=b{\int}_{0}^{1}{u}^{2n+1}{e}^{ux}{f}_{n}(x)\mathit{d}x=b{\left[{e}^{ux}{F}_{n}(x)\right]}_{0}^{1}=a{F}_{n}(1)b{F}_{n}(0).$$ 
Given the remarks on ${f}_{n}(x)$, ${w}_{n}$ should be an integer for all $n\in \mathbb{N}$, yet it is clear that $$ and so $\underset{n\to \mathrm{\infty}}{lim}{w}_{n}=0$, a contradiction^{}. ∎
The result could also easily have been obtained by starting with ${w}_{n}$ and integrating by parts $2n$ times. Note also that much stronger statements are known, such as “${e}^{a}$ is transcendental for all $a\in \mathbb{A}\setminus \{0\}$”^{2}^{2}$\mathbb{A}$ denotes the set of algebraic numbers^{}.. We conclude this entry with the following evident corollary:
Corollary.
For all $r\mathrm{\in}{\mathrm{Q}}^{\mathrm{+}}\mathrm{,}\mathrm{log}\mathit{}r$ is irrational.
References
 1 M. Aigner & G. M. Ziegler: Proofs from THE BOOK, 3${}^{\mathrm{rd}}$ edition (2004), SpringerVerlag, 30–31.
 2 G. H. Hardy & E. M. Wright: An Introduction to the Theory of Numbers, 5${}^{\mathrm{th}}$ edition (1979), Oxford University Press, 46–47.
Title  ${e}^{r}$ is irrational for $r\in \mathbb{Q}\setminus \{0\}$ 

Canonical name  ErIsIrrationalForRinmathbbQsetminus0 
Date of creation  20130322 15:07:46 
Last modified on  20130322 15:07:46 
Owner  Cosmin (8605) 
Last modified by  Cosmin (8605) 
Numerical id  12 
Author  Cosmin (8605) 
Entry type  Theorem 
Classification  msc 11J72 
Synonym  ${e}^{r}$ is irrational for nonzero rational r 
Synonym  irrationality of the exponential function^{} on $\mathbb{Q}$ 
Related topic  Irrational 
Related topic  EIsIrrationalProof 
Related topic  EIsIrrational 
Related topic  EIsTranscendental 