Definitions:

The natural numbers are the numbers specified by Peano's axioms, denoted by $\Bbb N$.

The integers are the numbers in $\Bbb N \times \Bbb N$ under the equivalence relationship $(a,b) = (c,d) \iff a+d = b+c$. A natural number $n$ is represented by the equivalence class $[(n,0)]$. They are denoted as $\Bbb Z$.

The rational numbers are the numbers in $\Bbb Z \times \Bbb Z^+$ under the equivalence relationship $(a,b) = (c,d) \iff ad = bc$. An integer $z$ is represented by the equivalence class $[(z,1)]$. They are denoted as $\Bbb Q$.

The real numbers are denoted as $\Bbb R$. They are defined in the following two ways:

1. Real numbers as Cauchy sequences: Let $(a_n)$ be a sequence of real numbers such that for any given $\epsilon > 0$, there is a natural number $N$ such that for all natural numbers which are greater than $m$ and $n$, $|a_m - a_n| < \epsilon$. Then, this sequence is a Cauchy sequence, and its limit is defined as a real number.

2. Real numbers as Dedekind cuts: a real number is defined a partition of the rational numbers into two sets, the smaller set containing no biggest element. The real number is the supremum of the smaller set (and the infimum of the larger set).

The complex numbers are the numbers in $\Bbb R \times \Bbb R$. A real number $r$ is represented by $(r,0)$.

Addition:

Addition in the natural numbers is defined recursively:

Addition in the integers is defined as $[(a,b)]+[(c,d)] = [(a+c,b+d)]$.

Addition in the rational numbers is defined as $[(a,b)] + [(c,d)] = [(ad+bc,bd)]$.

Addition in the real numbers as Cauchy sequences is defined as $[(x_n)] + [(y_n)] = [(x_n+y_n)]$.

Addition in the real numbers as Dedekind cuts is defined as $X+Y = \{x+y | x \in X, y\in Y\}$.

Addition in the complex numbers is defined as $(a,b)+(c,d) = (a+c,b+d)$.

Multiplication:

Multiplication in the natural numbers is defined recursively:

Multiplication in the integers is defined as $[(a,b)]\times[(c,d)] = [(ac+bd, ad+bc)]$.

Multiplication in the rational numbers is defined as $[(a,b)]\times[(c,d)] = [(ac,bd)]$.

Multiplication in the real numbers as Cauchy sequences is defined as $[(x_n)] \times [(y_n)] = [(x_n y_n)]$.

Multiplication in the real numbers as Dedekind cuts is too complicated and not quite relevant.

Multiplication in the complex numbers is defined as $(a,b) \times (c,d) = (ac-bd, ad+bc)$.

Polynomial:

Let $n$ be a natural number.

Let $x_1, \ldots, x_k$ be variables.

Construct the set $X = \displaystyle \bigcup_{i = 0}^n \left( \bigcup_{b_1,\cdots,b_k \in \Bbb N; b_1+\cdots+b_k = i} \{x_1^{b_1} x_2^{b_2} \cdots x_k^{b_k}\} \right)$.

Assign an element in $S$ for each element in $X$.

The sum of the elements in $X$ multiplied by the corresponding element in $S$ is a polynomial with coefficients in $S$ with degree $n$.

The zero polynomial is $f(x_1,\ldots,x_k) = 0$.

Linear independence:

Let $x_1, \ldots, x_k$ be complex numbers.

Let $a_1,\ldots,a_k$ be elements of a set $S$.

If $a_1 x_1 + \cdots + a_k x_k = 0 \implies a_1 = a_2 = \cdots = a_k = 0$, then $x_1,\ldots,x_k$ are linearly independent over $S$.

Algebraic independence:

Let $x_1,\ldots,x_k$ be complex numbers.

Let $f(z_1,\ldots,z_k)$ be a polynomial over a set $S$.

If $f(x_1,\ldots,x_k) = 0 \implies f(z_1,\ldots,z_k) = 0$, then $x_1,\ldots,x_k$ are algebraically independent over $S$.

Algebraic numbers are complex numbers roots of a polynomial with rational coefficients.

Transcendental numbers are complex numbers that are not algebraic.

Now the definitions have ended, and we can get to the more interesting parts.

Hermite-Lindemann-Weierstrass theorem (HLW):

If $a_1, \ldots, a_n$ are linearly independent over the rational numbers, then $e^{a_1}, \ldots, e^{a_n}$ are algebraically independent over the rational numbers.

Lindemann Theorem (L):

If $a$ is a non-zero algebraic number, then $e^a$ is transcendental.

Proof:

We use the case $n = 1$ in HLW.

That is, if $a$ is linearly independent over the rational numbers, then $e^a$ is algebraically independent over the rational numbers.

EDIT: the correct description of HLW is:

If $a_1, \ldots, a_n$ are algebraic numbers that are linearly independent over the rational numbers, then $e^{a_1}, \ldots, e^{a_n}$ are algebraically independent over the rational numbers.

EDIT: Corrected proof of L:

We use the case $n = 1$ in HLW.

That is, if $a$ is an algebraic number that is linearly independent over the rational numbers, then $e^a$ is algebraically independent over the rational numbers.

Which is the same as L.

Corollary to Lindemann Theorem (L/C):

If $a$ is an algebraic number that is neither $0$ nor $1$, then $\ln a$ is transcendental.

Proof:

Aiming for a contradiction, suppose that $\ln a$ is algebraic instead.

Since $a \ne 1$, $\ln a \ne 0$.

Therefore, by L, $e^{\ln a}$ is transcendental.

However, $e^{\ln a} = a$ which is assumed to be algebraic.

This is a contradiction.

Therefore, $\ln a$ must be transcendental.

Gelfond-Schneider Theorem (GS):

If $a$ and $b$ are algebraic numbers such that $a$ is neither zero nor one, and $b$ is irrational, then any value of $a^b$ is transcendental.

Conjectures:

Schanuel's Conjecture (S):

If $z_1, \ldots, z_n$ are complex numbers that are linearly independent, then the extension field $\Bbb (z_1, \ldots, z_n, e^{z_1}, \ldots, e^{z_n})$ has transcendence degree at least $n$ over the rational numbers.

Now we get to the interesting part:

Results:

1. $e$ is transcendental

Proof: use $a = 1$ in L.

By Euler's Identity, $e^{i\pi} = -1$.

Therefore, $\ln(-1) = i\pi$.

$-1$ is algebraic, as it is the root of the polynomial $x+1$.

Therefore, by L/C, $i \pi$ is transcendental.

Aiming for a contradiction, suppose that $\pi$ is algebraic.

Then, there exists a non-zero polynomial $f(x)$ with rational coefficients such that $f(\pi) = 0$.

Then, $g(x) := f(ix)f(-ix)$ would be a non-zero polynomial with rational coefficients such that $g(i\pi) = 0$.

Hence, $i\pi$ is algebraic, contradicting the fact that $i\pi$ is transcendental.

Therefore, $\pi$ must be transcendental.

3. if $a$ is a non-zero algebraic number, then $W(a)$ is transcendental

Here, $W$ is defined to be an inverse of $f(x) = xe^x$.

That is, $a = W(a)e^{W(a)}$.

The result is equivalent to that $a$ and $W(a)$ cannot be both algebraic.

In fact, if $a \ne 0$, then $W(a) \ne 0$.

Aiming for a contradiction, assume that $a$ and $W(a)$ are both algebraic.

By L, $e^{W(a)}$ is transcendental.

But it is equal to $a / W(a)$, which must be algebraic, because $a$ and $W(a)$ are both non-zero and algebraic.

(The algebraic numbers are closed under addition, subtraction, multiplication, and division, as long as the denominator is non-zero.)

Hence, a contradiction arises, and $W(a)$ must be transcendental.

4. $e^\pi$ is transcendental

By Euler's identity, $e^{i\pi} = -1$.

$-1$ is algebraic, because it is the root of the polynomial $x+1$.

$-i$ is algebraic, because it is a root of the polynomial $x^2+1$.

$-i$ is not rational, because it is not real.

Therefore, by GS, any value of $(-1)^{-i}$ is transcendental.

That is, $\left(e^{i\pi}\right)^{-i} = e^{-i^2\pi} = e^\pi$ is transcendental.