Proof of exponential law using limit definition of exponential function? For fun, I tried to prove the well-known exponential property $e^{a+b} = e^a e^b$ using the limit definition of the exponential function, below.

Definition. The exponential function is defined as follows.
$$e^x := \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon x \right)^{1/\epsilon}$$

I was able to outline the majority of the proof, however I do not have sufficient justification to go from line $\eqref 1$ to step $\eqref 2$. What limit properties might be used to fill in the blanks? I'd prefer not to use the binomial theorem or calculus-based argument, if possible (though if an expansion like that seems necessary, that is OK!).
Proof. 
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon (a+b) \right)^{1/\epsilon}$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon a + \epsilon b \right)^{1/\epsilon} \tag 1 \label 1$$
$$\vdots$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( 1 + \epsilon a + \epsilon b 
+ \epsilon^2 ab\right)^{1/\epsilon} \tag 2 \label 2$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left( (1 + \epsilon a)(1 + \epsilon b)\right)^{1/\epsilon}$$
$$e^{a+b} = \lim_{\epsilon \rightarrow 0} \left(1 + \epsilon a \right)^{1/\epsilon} \left(1 + \epsilon b\right)^{1/\epsilon}$$
$$e^{a+b} = e^a e^b$$
 A: Taking for granted that $e^x$ is well-defined in this way for all real $x$, it follows that for all real $x$:
$$
e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n.
$$
For all real $a$ and $b$, and every positive integer $n$,
\begin{align*}
& \phantom{={}} \left\lvert\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right\rvert \\
& = \frac{|ab|}{n^2}\left\lvert\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^{n-1} \!\! + \left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^{n-2}\left(1+\frac{a}{n}+\frac{b}{n}\right) + \cdots \right. \\
& \phantom{={}} \left. \cdots + \left(1+\frac{a}{n}+\frac{b}{n}\right)^{n-1}\right\rvert \\
& \leqslant  \frac{|ab|}{n}\left(1+\frac{|a|}{n}+\frac{|b|}{n}+\frac{|ab|}{n^2}\right)^{n-1}  = \frac{|ab|}{n}\left(1+\frac{|a|}{n}\right)^{n-1} \left(1+\frac{|b|}{n}\right)^{n-1} \\
& \leqslant \frac{|ab|}{n}\left(1+\frac{|a|}{n}\right)^n \left(1+\frac{|b|}{n}\right)^n,
\end{align*}
and this tends to zero as $n$ tends to infinity, because:
$$
\lim_{n\to\infty}\left(1+\frac{|a|}{n}\right)^n = e^{|a|}
\quad\text{and}\quad
\lim_{n\to\infty}\left(1+\frac{|b|}{n}\right)^n = e^{|b|}.
$$
Therefore:
\begin{align*}
e^ae^b & = \lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \\
& = \lim_{n\to\infty}\left[\left(\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right) + \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right] \\
& = \lim_{n\to\infty}\left(\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\right)^n \!\! - \left(1+\frac{a}{n}+\frac{b}{n}\right)^n\right) + \lim_{n\to\infty}\left(1+\frac{a}{n}+\frac{b}{n}\right)^n \\
& = 0 + e^{a+b} = e^{a+b}.
\end{align*}
