Constructing real exponents and related properties using solely $\lim_{n\to\infty }\left( 1+\dfrac{1}{n}\right) ^{n}$

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I'm trying to define $x^y$ for real $x$ and $y$, and also prove its related properties. The tools I have are

  • the definition $$e := \lim _{n\to\infty }\left( 1+\dfrac{1}{n}\right) ^{n}$$where $n \in \mathbb{N}$
  • Arithmetic operations over real numbers
  • Exponentiation to powers of rational numbers
  • Any other lemmas and theorems without involving $e^x$ or $\ln x$

I would like to generalize this definition to: $$ e^x := \lim _{n\to\infty }\left( 1+\dfrac{x}{n}\right) ^{n}$$for $x \in \mathbb{R}$, and then define $\ln x$ and $x^y := e^{y\ln x}$ so that I can prove $\dfrac{d}{dx}e^x = e^x$ and other properties. But first I have to show properties of indices like $$e^{x + y} = e^x e^y$$ for real $x$ and $y$.

In this case, using the definition for both $e^x$ and $e^y$ would yield $$e^x e^y = \lim_{n\rightarrow \infty }\left( 1+\dfrac{x+y}{n}+\dfrac{xy}{n^{2}}\right) ^{n}$$ How do I prove its equivalence to $\lim _{n\to\infty }\left( 1+\dfrac{x+y}{n}\right) ^{n}$ using only the tools mentioned above? Or, is there any better approaches for this property?

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Use Binomial Theorem. I'll let $1+\frac{x+y}{n}=a$, and $xy=b$.

$$a^n \le (a+\frac{b}{n^2})^n = \sum_{r=0}^n(\frac{b}{n^2})^r{a^{n-r}}_nC_r \le a^n+\sum_{r=1}^n{(\frac{b}{n^2})}^r{a^{n-r}}n^r = a^n+\sum_{r=1}^nb^r{a^{n-r}}n^{-r}$$ For sufficiently large $n$, $b^r{a^{n-r}}n^{-r} (0<r\le n)$ is ignorable.
So, following expression is reasonable. $$e^{x+y} = \lim _{n\to\infty }\left( 1+\dfrac{x+y}{n}\right) ^{n} = \lim _{n\to\infty }a^n = \lim _{n\to\infty }\left( a+\frac{b}{n^2}\right) ^{n} = \lim _{n\to\infty }\left( 1+\dfrac{x+y}{n}+\dfrac{xy}{n^2}\right) ^{n} = \lim _{n\to\infty }\left( 1+\dfrac{x}{n}\right) ^{n}\lim _{n\to\infty }\left( 1+\dfrac{y}{n}\right)^n = e^xe^y.$$