Derivative of Exponential Function/Proof 5/Lemma

Theorem
For $x \in\ R$, let $\left\lceil{x}\right\rceil$ denote the ceiling of $x$.

Then:
 * $\forall x \in \R : n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing.

Proof
First:

Then:

So, for $n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil$:
 * $\left\langle{\dfrac n {n + x} }\right\rangle$

and:
 * $\left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$

are positive.

Now let $n \ge \left\lceil{\left\vert{x}\right\vert}\right\rceil$.

Suppose first that $x \in \R_{\ge 0}$.

Then:

So $\left\langle{\dfrac n {n + x} }\right\rangle$ is increasing.

Further, from Exponential Sequence is Eventually Increasing:
 * $\left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing.

Hence, from Product of Positive Increasing Functions is Increasing:
 * $n \ge \left\lceil{\left\vert{x}\right\vert }\right\rceil \implies \left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is increasing.

Suppose instead that $x \in \R_{<0}$.

that:
 * $\left\langle{\dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle$ is decreasing.

From above: $\left\langle{1 + \dfrac x n}\right\rangle = \left\langle{\dfrac {n + x} n}\right\rangle$ is decreasing.

Thus, from Product of Positive Increasing and Decreasing Functions is Decreasing:
 * $\left\langle{\dfrac {n + x} n \dfrac n {n + x} \left({1 + \dfrac x n}\right)^n}\right\rangle = \left\langle{\left({1 + \dfrac x n}\right)^n}\right\rangle$ is decreasing.

This contradicts Exponential Sequence is Eventually Increasing.

Hence the result, by Proof by Contradiction.