Exponential Sequence is Eventually Increasing

Theorem
Let $\left \langle{E_n}\right \rangle$ be the sequence of  mappings $E_n: \R \to \R$ defined as:
 * $E_n(x) = \left({1 + \dfrac{x}{n}}\right)^n$.

Then, for sufficiently large $n \in \N$, $\left \langle{E_n(x)}\right \rangle$ is increasing with respect to $n$.

That is:
 * $\forall x \in \R : \forall n \in \N : n \geq \left\lceil{ \left\vert{ x }\right\vert }\right\rceil \implies E_n \left({ x }\right) \leq E_{n+1} \left({ x }\right)$

where $\left \lceil {\cdot} \right \rceil$ denotes the ceiling function.

Corollary

 * $\forall x \in \R : \forall n \in \N : n \geq \left\lceil{ \left\vert{ x }\right\vert }\right\rceil \implies E_n \left({ x }\right) > 0$

Proof
Fix $x \in \R \setminus \left \{ {0} \right \}$.

Then:

This proves the corollary.

Also we may apply the AM-GM inequality, with $x_1 := 1$ and $x_2 := \ldots := x_{n+1} = 1 + \dfrac{x}{n}$, to obtain that:
 * $\dfrac{ 1 + n \left({ 1 + \dfrac{x}{n} }\right) }{n + 1} > \left({ \left({ 1 + \dfrac{x}{n} }\right) ^n }\right) ^{ 1/ \left({ n+1 }\right) }$

Or, after some simplification:
 * $1 + \dfrac{x}{n+1} > \left({ 1 + \dfrac{x}{n} }\right)^{ n / \left({ n+1 }\right) }$

Finally, from Power Function is Strictly Increasing over Positive Reals: Natural Exponent:
 * $\left({ 1 + \dfrac{x}{n+1} }\right)^{n+1} > \left({ 1 + \dfrac{x}{n} }\right)^{n}$

For $x = 0$:
 * $\forall n \in \N : E_n \left({ 0 }\right) = 1$

The result follows from Mapping Constant iff Increasing and Decreasing.