Theorem: Exponential of Sum/Real Numbers

Let $x, y \in \R$ be real numbers.
Let $\exp x$ be the exponential of $x$.
Then:
:$\map \exp {x + y} = \paren {\exp x} \paren {\exp y}$

Proof:
This proof assumes the definition of $\exp$ as defined by a limit of a sequence:
:$\exp x = \ds \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n$
From Powers of Group Elements we can presuppose the Exponent Combination Laws for natural number indices.
By definition:
{{begin-eqn}}
{{eqn | l = \paren {\exp x} \paren {\exp y}
      | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac x n}^n \lim_{n \mathop \to +\infty} \paren {1 + \frac y n}^n
}}
{{eqn | r = \lim_{n \mathop \to +\infty} \paren {\paren {1 + \frac x n} \paren {1 + \frac y n} }^n
      | c = Combination Theorem for Limits of Real Functions
}}
{{eqn | r = \lim_{n \mathop \to +\infty} \paren {1 + \frac{x + y} n + \frac{x y} {n^2} }^n
}}
{{end-eqn}}
Intuitively, the $\paren {1 + \dfrac {x + y} n}$ term is the most influential of the terms involved in the limit, and:
:$\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n \to \paren {1 + \dfrac {x + y} n}^n$ as $n \to +\infty$
To formalize this claim:
:$\map \exp {x + y} = \exp x \cdot \exp y \iff \dfrac {\exp x \cdot \exp y} {\map \exp {x + y} } = 1$
{{begin-eqn}}
{{eqn | l = \frac {\paren {1 + \frac {x + y} n + \frac {x y} {n^2} }^n} {\paren {1 + \frac{x + y} n}^n}
      | r = \paren {1 + \frac{x y} {n^2 + n x + n y} }^n
      | c = Lemma
}}
{{eqn | r = \sum_{k \mathop = 0}^n \binom n k \paren {\frac {x y} {n^2 + n x + n y} }^k
      | c = Binomial Theorem
}}
{{eqn | r = 1 + \sum_{k \mathop = 1}^n \binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k
}}
{{end-eqn}}
Now, as $n \to +\infty$, we use the Combination Theorem for Limits of Real Functions to investigate the behavior of this sequence, term by term.
As $1$ trivially converges to $1$, consider now the other terms of the sequence.
We invoke the Squeeze Theorem for Absolutely Convergent Series.
Hence it will suffice to investigate the limit behaviour of:
:$\ds \sum_{k \mathop = 1}^n \, \size {\binom n k n^{-k} \paren {\frac {x y} {n + x + y} }^k}$
From $\dbinom n k$ is not greater than $n^k$:
{{begin-eqn}}
{{eqn | l = \binom n k
      | o = \le
      | r = n^k
      | c = for all $n, k$ here considered
}}
{{eqn | ll= \leadsto
      | l = \binom n k n^{-k} 
      | o = \le
      | r = 1
      | c = divide both sides by $n^k$
}}
{{end-eqn}}
Therefore, we may conclude, using Absolute Value is Bounded Below by Zero:
:$\ds 0 \le \sum_{k \mathop = 1}^n \size { {n \choose k} n^{-k} \paren {\frac {x y} {n + x + y} }^k} \le \sum_{k \mathop = 1}^n \size {\frac {x y} {n + x + y} }^k$
From Sum of Infinite Geometric Sequence, the right hand term converges to:
{{begin-eqn}}
{{eqn | l = \frac {\dfrac {x y} {n + x + y} } {1 - \dfrac {x y} {n + x + y} }
      | r = \frac {x y} {n + x + y - x y}
}}
{{eqn | o = \to
      | r = 0
      | c = as $n \to +\infty$
}}
{{end-eqn}}
:$0 \to 0$ as $n \to +\infty$, trivially.
This means:
:$\dfrac {\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n} {\paren {1 + \dfrac {x + y} n}^n} \to 1$ as $n \to +\infty$
which is equivalent to our hypothesis:
:$\paren {1 + \dfrac {x + y} n + \dfrac {x y} {n^2} }^n \to \paren {1 + \dfrac {x + y} n}^n$ as $n \to +\infty$
{{qed}}