Proof that $\exp(x+y) = \exp(x)*\exp(y)$ using limit definition of $\exp(x)$ I want to prove: $\exp(x+y) = \exp(x)\cdot \exp(y)$ using the definition: $\exp(x) = \lim_{n\to\infty} (1+\frac{x}{n})^n$ 
I am having trouble completing the proof, but here is my idea so far: $$\lim_{n\to\infty} \left(1+\frac{x+y}{n}\right)^n = \lim_{n\to\infty} \left(1+\frac{x}{n}\right)^n \cdot \lim_{n\to\infty} \left(1+\frac{y}{n}\right)^n =   
 \lim_{n\to\infty} \left(\left(1+\frac{x}{n}\right)^n \cdot \left(1+\frac{y}{n}\right)^n  \right) $$
Now I rearrange the last expression: $$\lim_{n\to\infty} \left(1+\frac{x+y+\frac{xy}{n}}{n}\right)^n $$
From here my idea is to somehow show that this limit is equal to $$\lim_{n\to\infty} \left(1+\frac{x+y}{n}\right)^n = \exp(x+y)$$ using the Squeeze Theorem and perhaps Bernoulli's Inequality, but I am at a loss as to how exactly to do it. I'd appreciate your help. 
 A: It suffices to show that
$$
\lim_{n\to\infty}\frac{\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n}{\left(1+\frac{x+y}{n}\right)^n}=1 \tag{1}
$$
But
$$
\frac{\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n}{\left(1+\frac{x+y}{n}\right)^n}=
\left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^n
$$
and as, for suitably large $n$, say $n\ge n_0$, we have that 
$$
\frac{1}{2}<1+\frac{x+y}{n}<2,
$$ 
then 
$$
\left(1+\frac{xy}{2n^2}\right)^n<\frac{\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n}{\left(1+\frac{x+y}{n}\right)^n}=
\left(1+\frac{xy}{n^2(1+\frac{x+y}{n})}\right)^n<\left(1+\frac{2xy}{n^2}\right)^n
$$
Now, for all $z\in\mathbb R$, we have
$$
\left(1+\frac{z}{n^2}\right)^n=\left(\left(1+\frac{z}{n^2}\right)^{n^2}\right)^{1/n}\to 1,
$$
since $\left(1+\frac{z}{n^2}\right)^{n^2}\to e^z$.
Thus
$$
\lim_{n\to\infty}\left(1+\frac{xy}{2n^2}\right)^n=\lim_{n\to\infty}\left(1+\frac{2xy}{n^2}\right)^n=1,
$$
and hence $(1)$ holds.
A: $$
\\\dfrac{e^{x+y}}{e^x}=\lim_{n\to+\infty}{\Big(\dfrac{1+\frac {x+y} n}{1+\frac x n}\Big)^n}=
\\\lim_{n\to+\infty}\Big(\frac{x+y+n}{x+n}\Big)^n=\lim_{n\to+\infty}\Big(1+\frac y{x+n}\Big)^n=
\\\lim_{n\to+\infty}\frac{\Big(1+\frac y {x+n}\Big)^{n+x}}{\Big(1+\frac y {x+n}\Big)^x}=e^y
$$
Because $\lim_{n\to+\infty}{\Big(1+\frac y {x+n}\Big)^{n+x}}=e^y$ and $\lim_{n\to+\infty}{\Big(1+\frac y {x+n}\Big)^x}=1$
