Show identity for $e^x$ I want to show that, using the definition $e^x := \lim_{n \to \infty} (1+\frac{x}{n})^n$, the following identity holds:
$$ e^{x+y}=e^xe^y$$
This is what I tried:
\begin{align}
e^xe^y &=\lim_{n \to \infty} \left(1+\frac{x}{n}\right)^n\cdot\left(1+\frac{y}{n}\right)^n \\
&= \lim_{n \to \infty}\left(\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right)\right)^n \\
&=\lim_{n \to \infty}\left(1+\frac{x+y}{n}+\frac{xy}{n^2}\right)^n
\end{align}
However, this seems to be more than $e^{x+y}=\lim_{n \to \infty}(1+\frac{x+y}{n})^n$. How can I resolve this seeming contradiction to the upper identity?
 A: For $x, y \geq 0$ we have:
$$\left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) = 1 + \frac{x+y}{n} + \frac{xy}{n^2}\geq 1+\frac{x+y}{n}$$
For $n \in \mathbb{N}$, using the identity $u^n-v^n = (u-v)\sum_{k=0}^{n-1}u^{n-1-k}v^k$ we have:
\begin{align}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n - \left(1 + \frac{x+y}{n}\right)^n &= \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^n - \left(1+\frac{x+y}{n}\right)^n \\
&= \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k
\end{align}
Using the first inequality we get:
$$\frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1} \leq \left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n  - \left(1+\frac{x+y}{n}\right)^n \leq \frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1}$$
The left inequality is obtained by noticing that $\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right) = \left(1+\frac{x}{n}\right)\left(1+\frac{y}{n}\right) \geq \left(1+\frac{x+y}{n}\right)$ so by replacing the larger term with the smaller one in every product, we get a lower bound for the sum:
\begin{align}\frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k &\geq \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1+\frac{x+y}{n}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k \\
&= \frac{xy}{n^2}\sum_{k=0}^{n-1}\left(1+\frac{x+y}{n}\right)^{n-1} \\
&= \frac{xy}{n^2} \cdot n\,\left(1+\frac{x+y}{n}\right)^{n-1} \\
&= \frac{xy}{n} \left(1+\frac{x+y}{n}\right)^{n-1}
\end{align}
Similarly, for the right inequality we replace the smaller term with the larger one:
\begin{align}\frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1+\frac{x+y}{n}\right)^k &\leq \frac{xy}{n^2}\sum_{k=0}^{n-1} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1-k}\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^k \\
&= \frac{xy}{n^2}\sum_{k=0}^{n-1}\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\
&= \frac{xy}{n^2} \cdot n\,\left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\
&= \frac{xy}{n} \left(1 + \frac{x+y}{n} + \frac{xy}{n^2}\right)^{n-1} \\
&= \frac{xy}{n} \left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1}
\end{align}
Letting $n\to\infty$ both sides converge to $0$ so by the squeeze theorem we have:
$$\lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n  - \left(1+\frac{x+y}{n}\right)^n = 0$$
So
$$e^xe^y = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n\left(1+\frac{y}{n}\right)^n = \lim_{n\to\infty}\left(1+\frac{x+y}{n}\right)^n = e^{x+y}$$
Just for clarification, the limits of the lower and upper bound are obtained using the limit of product and quotient theorem:
$$\lim_{n\to\infty} \frac{xy}{n}\left(1+\frac{x+y}{n}\right)^{n-1} = \lim_{n\to\infty} \frac{xy}{n} \frac{\left(1+\frac{x+y}{n}\right)^n}{\left(1+\frac{x+y}{n}\right)} = 0 \cdot\frac{e^{x+y}}{1} = 0$$
$$\lim_{n\to\infty} \frac{xy}{n}\left(1+\frac{x}{n}\right)^{n-1}\left(1+\frac{y}{n}\right)^{n-1} = \lim_{n\to\infty}\frac{xy}{n}\frac{\left(1+\frac{x}{n}\right)^{n-1}}{\left(1+\frac{x}{n}\right)}\frac{\left(1+\frac{y}{n}\right)^{n-1}}{\left(1+\frac{y}{n}\right)} = 0\cdot \frac{e^x}{1}\cdot\frac{e^y}{1} = 0$$
