I am trying to derive the power series representation for $e^x$ for rational exponents without using Taylor series. Defining $e$ as $$e:=\lim_{n(\in\mathbb{N})\rightarrow\infty}\left(1+\frac{1}{n}\right)^n,$$ I want to show that $e^x=\left( \lim_{n(\in\mathbb{N})\rightarrow\infty}\left(1+\frac{1}{n}\right)^n \right)^x= \lim_{n(\in\mathbb{N})\rightarrow\infty}\left(1+\frac{1}{n}\right)^{nx}$ is equivalent to $\sum_0^\infty\frac{x^k}{k!}$ for all rational exponents. Using the ratio test, I know that this power series is convergent for all real (and thus rational) $x$; I just need to prove that it converges to $e^x$.
For $x\in \mathbb{N}$, $nx\in \mathbb{N}$, so I can just use $t=nx$, and use the (regular) binomial theorem to expand $(1+\frac{x}{t})^{t} $ inside the limit and then take it to arrive at the power series. I'm not sure what to do with non-natural exponents however. I know I can use Newton's Generalized Binomial Theorem, but all proofs of that involve Taylor series or differential equations, which defeats the purpose.
$$\tag{1}e=\sum^\infty_{k=0}\frac{1}{k!}\text{.}$$ By induction and Mertens' theorem, we get, for each positive integer $n$, $$\tag{2}e^n=\sum^\infty_{k=0}\frac{n^k}{k!}\text{.}$$ For $$e^{n+1}=e^ne=\sum^\infty_{k=0}\frac{n^k}{k!}\sum^\infty_{k=0}\frac{1}{k!}=\sum^\infty_{k=0}\sum^k_{m=0}\frac{n^m}{m!}\frac{1}{(k-m)!}=\sum^\infty_{k=0}\frac{(n+1)^k}{k!}\text{.}$$ Similarly, for each positive integer $n$, we have $$\left(\sum^\infty_{k=0}\frac{(\frac{1}{n})^k}{k!}\right)^n=\sum^\infty_{k=0}\frac{1}{k!}=e\text{,}$$ so that $$\tag{3}e^{1/n}=\sum^\infty_{k=0}\frac{(\frac{1}{n})^k}{k!}\text{.}$$ Again, by the theorem about Cauchy product, we have $$\tag{4}\left(\sum^\infty_{k=0}\frac{(\frac{m}{n})^k}{k!}\right)^n=e^m$$ for any positive integers $m$, $n$, and $$\tag{5}e^p\sum^\infty_{k=0}\frac{(-p)^k}{k!}=1$$ for any positive rational $p$. Hence $$\tag{6}e^p=\sum^\infty_{k=0}\frac{p^k}{k!}$$ for all rational $p$.
Is this what you want?