We will handle $x>0$ here.
If we define $e=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^n$, then $e^x=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}$. Note that since $0\le nx-\lfloor nx\rfloor<1$, $ \begin{align} e^x&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor} \left(1+\frac{1}{n}\right)^{nx-\lfloor nx\rfloor}\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor} \lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx-\lfloor nx\rfloor}\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor} \end{align} $ Using the binomial theorem, $ \begin{align} \left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor} &=\sum_{k=0}^{\lfloor nx\rfloor} \frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}\\ &=\sum_{k=0}^\infty \frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k} \end{align} $ Where $P(n,k)=n(n-1)(n-2)...(n-k+1)$ is the number of permutations of $n$ things taken $k$ at a time.
Note that $0\le\frac{P({\lfloor nx\rfloor},k)}{n^k}\le x^k$ and that $\sum_{k=0}^\infty \frac{x^k}{k!}$ converges absolutely. Thus, if we choose an $\epsilon>0$, we can find an $N$ large enough so that, for all $n$, $ 0\le\sum_{k=N}^\infty \frac{1}{k!}\left(x^k-\frac{P({\lfloor nx\rfloor},k)}{n^k}\right)\le\frac{\epsilon}{2} $ Furthermore, note that $\lim_{n\to\infty}\frac{P({\lfloor nx\rfloor},k)}{n^k}=x^k$. Therefore, we can choose an $n$ large enough so that $ 0\le\sum_{k=0}^{N-1} \frac{1}{k!}\left(x^k-\frac{P({\lfloor nx\rfloor},k)}{n^k}\right)\le\frac{\epsilon}{2} $ Thus, for n large enough, $ 0\le\sum_{k=0}^\infty \frac{1}{k!}\left(x^k-\frac{P({\lfloor nx\rfloor},k)}{n^k}\right)\le\epsilon $ Therefore, $ \lim_{n\to\infty}\;\sum_{k=0}^\infty\frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}=\sum_{k=0}^\infty\frac{x^k}{k!} $ Summarizing, we have $ \begin{align} e^x&=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{nx}\\ &=\lim_{n\to\infty}\left(1+\frac{1}{n}\right)^{\lfloor nx\rfloor}\\ &=\lim_{n\to\infty}\;\sum_{k=0}^\infty \frac{1}{k!}\frac{P({\lfloor nx\rfloor},k)}{n^k}\\ &=\sum_{k=0}^\infty\frac{x^k}{k!} \end{align} $