Euler's Number is Transcendental/Proof 1

Proof
there exist integers $a_0, \ldots, a_n$ with $a_0 \ne 0$ such that:


 * $(1): \quad a_n e^n + a_{n - 1} e^{n - 1} + \cdots + a_0 = 0$

Define $M$, $M_1, \ldots, M_n$ and $\epsilon_1, \ldots, \epsilon_n$ as follows:

where $p$ is a prime number with $p > n$ and $p > \size {a_0}$.

The expression $\sqbrk {\paren {x - 1} \cdots \paren {x - n} }^p$ is a polynomial of degree $n p$ with integer coefficients.

Hence:

where $C_\alpha$ are integers and $C_0 = \pm \paren {n!}^p$.

For $\alpha = 0$ we have:


 * $C_0 \dfrac {\paren {p - 1}!} {\paren {p - 1}!} = \pm \paren {n!}^p$

Since $p > n$, it follows from Prime iff Coprime to all Smaller Positive Integers and Euclid's Lemma that this term is not divisible by $p$.

For $\alpha \ge 1$ we have:


 * $C_\alpha \dfrac {\paren {p - 1 + \alpha}!} {\paren {p - 1}!} = C_\alpha \paren {p - 1 + \alpha} \paren {p - 2 + \alpha} \cdots p$

which is clearly divisible by $p$.

It follows from Common Divisor Divides Difference that $M$ is an integer not divisible by $p$.

We also have:

The expression $\sqbrk {\paren {x + k - 1} \cdots \paren {x + k - n} }$ is divisible by $x$.

So $\paren {x + k}^{p - 1} \sqbrk {\paren {x + k - 1} \cdots \paren {x + k - n} }^p$ is a polynomial of degree at least $p$ with integer coefficients.

Hence:

where $D_\alpha$ are integers.

Since this sum begins with $\alpha = 1$, each term is divisible by $p$.

Thus each $M_k$ is an integer divisible by $p$.

By the above definitions we have:


 * $e^k = \dfrac {M_k + \epsilon_k} M$

Substituting this into $(1)$ and multiplying by $M$ we obtain:


 * $\paren {a_0 M + a_1 M_1 + \cdots + a_n M_n} + \paren {a_1 \epsilon_1 + \cdots + a_n \epsilon_n} = 0$

Since $p > \size {a_0}$, it follows from Prime iff Coprime to all Smaller Positive Integers that $p$ does not divide $a_0$.

So by Euclid's Lemma, $a_0 M$ is not divisible by $p$.

Since each $M_k$ is divisible by $p$, it follows from Common Divisor Divides Difference that $a_0 M + a_1 M_1 + \cdots + a_n M_n$ is not divisible by $p$.

Therefore $a_0 M + a_1 M_1 + \cdots + a_n M_n$ is a non-zero integer.

We also have:

Let $A$ be the maximum value of $\size {\paren {x - 1} \cdots \paren {x - n} }$ for $x$ in the interval $\closedint 0 n$.

Then:

By Power over Factorial:


 * $\displaystyle \lim_{p \mathop \to \infty} \frac {e^n \paren {n A}^p} {p!} = 0$

So $\size {\epsilon_k}$, and therefore $\size {a_1 \epsilon_1 + \cdots + a_n \epsilon_n}$, can be made arbitrarily small by choosing $p$ sufficiently large.

It follows that $\paren {a_0 M + a_1 M_1 + \cdots + a_n M_n} + \paren {a_1 \epsilon_1 + \cdots + a_n \epsilon_n}$ is non-zero.

This is a contradiction, so $e$ must be transcendental.