Exponential of Sum/Real Numbers/Lemma

Lemma
Let $x, y \in \R$.

Let $n \in \N_{> 0}$ such that $n > -\paren {x + y}$.

Then:


 * $1 + \dfrac {x + y} n + \dfrac {x y} {n^2} = \paren {1 + \dfrac {x + y} n} \paren {1 + \dfrac {\paren {\frac {x y} {n + x + y} } } n}$

Proof
As $n \in \N_{> 0}$ we have that $n \ne 0$ and so the fractions in the expressions are defined.

That final step is justified, as we have that $n > -\paren {x + y}$ and so $n + x + y \ne 0$.

Also see
This lemma is used in:
 * Proof 2 of (Real) Exponential of Sum
 * Proof 3 of (Real) Exponential of Sum


 * $\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$

for real $x$ and $y$.