The answer from robjohn linked in his comment is a must visit. It deals with the definition of $e^{x}$ as $\lim_{n \to \infty}(1 + 1/n)^{nx}$. Another direct approach starting with $(1 + x/n)^{n}$ is given below.
Let $$F_{n}(x) = 1 + x + \frac{x^{2}}{2!} + \cdots + \frac{x^{n}}{n!}$$ so that $\lim_{n \to \infty}F_{n}(x) = F(x)$ exist and $F(x)$ is given by the convergent power series $$F(x) = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots = \sum_{k = 0}^{\infty}\frac{x^{k}}{k!}$$ First we assume that $x > 0$. We have to establish that $\lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} = F(x)$
If $n > x$ then we have $$\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \frac{n(n - 1)}{2!}\left(\frac{x}{n}\right)^{2} + \cdots + \frac{n!}{n!}\left(\frac{x}{n}\right)^{n}$$ so that $$\left(1 + \frac{x}{n}\right)^{n} < F_{n}(x)$$
Next consider the expression $(1 - (x/n))^{-n}$. By the binomial theorem for negative exponents we have $$\left(1 - \frac{x}{n}\right)^{-n} = 1 + x + \frac{n(n + 1)}{2!}\left(\frac{x}{n}\right)^{2} + \cdots$$ so that $$\left(1 - \frac{x}{n}\right)^{-n} \geq F_{n}(x)$$ It follows that for $0 < x < n$ we have $$\left(1 + \frac{x}{n}\right)^{n} < F_{n}(x) \leq \left(1 - \frac{x}{n}\right)^{-n}\,\,\,\cdots (1)$$
Now note that $\lim_{n \to \infty}(1 + x/n)^{n} = G(x)$ exists and we have $(1 + (x/n))^{n} < G(x)$ for all $x > 0$ and all positive integers $n$. Therefore
$\displaystyle \begin{aligned}\left(1 - \frac{x}{n}\right)^{-n} - \left(1 + \frac{x}{n}\right)^{n} &= \left(1 + \frac{x}{n}\right)^{n}\left\{\left(1 - \frac{x^{2}}{n^{2}}\right)^{-n} - 1\right\}\\
&< G(x)\left\{\left(1 - \frac{x^{2}}{n}\right)^{-1} - 1\right\} = \frac{x^{2}G(x)}{n - x^{2}}\end{aligned}$
so that the expression $(1 - (x/n))^{-n} - (1 + (x/n))^{n}$ tends to $0$ as $n \to \infty$ (using squeeze theorem). It follows that $$\lim\limits_{n \to \infty}\left(1 - \dfrac{x}{n}\right)^{-n} = \lim\limits_{n \to \infty}\left(1 + \dfrac{x}{n}\right)^{n} = G(x)\,\,\,\cdots (2)$$
Now applying squeeze theorem on equation $(1)$ and using equation $(2)$ above we get $\lim_{n \to \infty}F_{n}(x) = G(x)$ so that $F(x) = G(x)$ for all $x > 0$.
For $x = 0$ the result is trivial. For $x < 0$ we need to use the fact that $F(x + y) = F(x)F(y)$ (this is done by multiplication of series) and thus $F(-x) = 1/F(x)$. Similarly from equation $(2)$ it follows that $G(-x) = 1/G(x)$. Thus the relation $F(x) = G(x)$ holds for every real $x$. Noting the definitions of $F, G$ we see that $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = 1 + x + \frac{x^{2}}{2!} + \frac{x^{3}}{3!} + \cdots $$ for all $x \in \mathbb{R}$.
Update: While providing an answer to another question I came up with an alternative proof of $$\lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n} = \lim_{n \to \infty}\left(1 - \frac{x}{n}\right)^{-n}$$ which is worth mentioning here.
Let us set $$f(n) = \left(1 + \frac{x}{n}\right)^{n}, g(n) = \left(1 - \frac{x}{n}\right)^{-n}$$ and as in the earlier proof we only need to handle the case for $x > 0$. By the use of binomial theorem we can show that the sequence $f(n)$ increases and is bounded by $F(x)$ so that $f(n)$ tends to a limit as $n \to \infty$ and the limit is greater than or equal to $(1 + x)$.
Now we can see that if $0 < x^{2} < n$ then $$\begin{aligned}\frac{f(n)}{g(n)} &= \left(1 - \frac{x^{2}}{n^{2}}\right)^{n}\\
&= 1 - n\cdot\frac{x^{2}}{n^{2}} + \frac{n(n - 1)}{2!}\left(\frac{x^{2}}{n^{2}}\right)^{2} - \cdots + \\
&= 1 - \frac{x^{2}}{n} + \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} - \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&= 1 - \phi(n)\end{aligned}$$ where $\phi(n)$ is a finite sum defined by $$\phi(n) = \frac{x^{2}}{n} - \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} - \cdots$$ We can estimate $\phi(n)$ as follows $$\begin{aligned}0 \leq |\phi(n)| &\leq \frac{x^{2}}{n} + \dfrac{\left(1 - \dfrac{1}{n}\right)}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \dfrac{\left(1 - \dfrac{1}{n}\right)\left(1 - \dfrac{2}{n}\right)}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&\leq \frac{x^{2}}{n} + \frac{1}{2!}\left(\frac{x^{2}}{n}\right)^{2} + \frac{1}{3!}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&\leq \frac{x^{2}}{n} + \frac{1}{2}\left(\frac{x^{2}}{n}\right)^{2} + \frac{1}{2^{2}}\left(\frac{x^{2}}{n}\right)^{3} + \cdots\\
&= \frac{x^{2}}{n}\left\{1 + \frac{x^{2}}{2n} + \left(\frac{x^{2}}{2n}\right)^{2} + \cdots\right\}\\
&= \frac{x^{2}}{n}\cdot\dfrac{1 - \left(\dfrac{x^{2}}{2n}\right)^{n}}{1 - \dfrac{x^{2}}{2n}}\\
&\leq \frac{x^{2}}{n}\cdot\dfrac{1}{1 - \dfrac{x^{2}}{2n}}\to 0 \text{ as }n \to \infty\end{aligned}$$ and then we see that $\phi(n) \to 0$ as $n \to \infty$. It follows that $f(n)/g(n) \to 1$ as $n \to \infty$ and therefore $\lim_{n \to \infty}f(n) = \lim_{n \to \infty}g(n)$.
I'm not convinced. The "stuff" consists of various small terms that all will become very small, but at the same time, there will be more and more of these terms as $n \to \infty$. As with a Riemann sum, we expect to sum a lot of small quantities and take the limit, but this doesn't imply the Riemann integral is always $0$.
Here's how I would approach it:
Fix $\varepsilon > 0$, and let $N \in \Bbb{N}$ be such that
$$n \ge N \implies |x_n - x| < \varepsilon$$
and $x_n > -N$ (which we can do, as convergent sequences are bounded). Then, for $n > N$,
$$\prod_{k=1}^n \left(1 + \frac{x_k}{n}\right) = \prod_{k=1}^{N-1} \left(1 + \frac{x_k}{n}\right) \cdot \prod_{k=N}^{n} \left(1 + \frac{x_k}{n}\right),$$
where
$$\left(1 + \frac{x - \varepsilon}{n}\right)^{n-N+1} \le \prod_{k=N}^{n} \left(1 + \frac{x_k}{n}\right) \le \left(1 + \frac{x + \varepsilon}{n}\right)^{n-N+1}.$$
Since $x_n > -N$ and $n > N$, we have
$$\prod_{k=1}^{N-1} \left(1 + \frac{x_k}{n}\right) > 0,$$
so
$$\left(1 + \frac{x - \varepsilon}{n}\right)^{n-N+1} \prod_{k=1}^{N-1} \left(1 + \frac{x_n}{n}\right) \le \prod_{k=1}^{n} \left(1 + \frac{x_k}{n}\right) \le \left(1 + \frac{x + \varepsilon}{n}\right)^{n-N+1} \prod_{k=1}^{N-1} \left(1 + \frac{x_n}{n}\right).$$
Taking the limit as $n \to \infty$,
$$e^{x - \varepsilon} \le \prod_{k=1}^{\infty} \left(1 + \frac{x_k}{n}\right) \le e^{x + \varepsilon},$$
for all $\varepsilon > 0$. So, if you accept that the product converges, the limit must be $e^x$, by the continuity of $e^x$.
Best Answer
You do not need to deal with that hard expansion. Here is another way using Squeeze theorem for $xy\ge 0$. The case $xy\le 0$ is also similar. Notice that for any $a>0$ and large enough $n$ we have:$$1+{x+y\over n}\le 1+{x+y\over n}+{xy\over n^2}<1+{x+y\over n}+{xy\over an}=1+{x+y+{xy\over a}\over n}$$using Squeeze theorem we have$$\lim _{n\to \infty}(1+{x+y\over n})^n\le \lim _{n\to \infty}(1+{x+y\over n}+{xy\over n^2})^n\le \lim _{n\to \infty}(1+{x+y+{xy\over a}\over n})^n$$or using the definition $$E(x+y)\le E(x)E(y)\le E(x+y+{xy\over a})$$since this is true for any $a>0$ by tending $a$ to $\infty$ we obtain $$E(x+y)\le E(x)E(y)\le E(x+y)$$which yields to $$E(x+y)=E(x)E(y)$$