Exponential Sequence is Eventually Increasing

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

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

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

where $\left \lceil {x} \right \rceil$ denotes the ceiling of $x$.

Proof
Fix $x \in \R$.

Then:

So 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:


 * $\dfrac{1 + n \left({1 + \dfrac x n}\right)} {n + 1} > \left({\left({1 + \dfrac x n}\right)^n}\right)^{1 / \left({n + 1}\right)}$

After simplification:
 * $1 + \dfrac x {n + 1} > \left({1 + \dfrac x n}\right)^{n / \left({n + 1}\right)}$

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

where we have raised both sides to the power of $n + 1$.

Hence the result.