Exponential of Rational Number is Irrational

Theorem
Let $r$ be a rational number such that $r \ne 0$.

Then:
 * $e^r$ is irrational

where $e$ is Euler's number.

Proof
Let $r = \dfrac p q$ be rational such that $r \ne 0$.

$e^r$ is rational.

Then $\left({e^r}\right)^q = e^p$ is also rational.

Then if $e^{-p}$ is rational, it follows that $e^p$ is rational.

It is therefore sufficient to derive a contradiction from the supposition that $e^p$ is rational for every $p \in \Z_{>0}$.

Let $e^p = \dfrac a b$ for $a, b \in \Z_{>0}$.

Let $n \in \Z_{>0}$ be a strictly positive integer.

Let $f \left({x}\right)$ be the function defined as:

Note that:
 * $(2): \quad 0 < x < 1 \implies 0 < f \left({x}\right) < \dfrac 1 {n!}$

We also have:
 * $f \left({0}\right) = 0$


 * $f^{\left({m}\right)} \left({0}\right) = 0$ if $m < n$ or $m > 2 n$

and:
 * $n \le m \le 2 n \implies f^{\left({m}\right)} \left({0}\right) = \dfrac {m!} {n!} c_m$

and this number is an integer.

Thus at $x = 0$, $f \left({x}\right)$ and all its derivatives are integers.

Since $f \left({1 - x}\right) = f \left({x}\right)$, the same is true for $x = 1$.

Let $F \left({x}\right)$ be the function defined as:

Because of the properties of $f \left({x}\right)$ and its derivatives above, $F \left({0}\right)$ and $F \left({1}\right)$ are integers.

Next we have:

$(4)$ leads to:

which is an integer.

But from $(2)$:

The expression on the tends to $0$ as $n$ tends to $\infty$.

Hence:
 * $0 < a F \left({1}\right) - b F \left({0}\right) < 1$

for sufficiently large $n$.

But there exists no integer strictly between $0$ and $1$.

From this contradiction it follows that our original assumption, that is, that $e^r$ is rational, must have been false.