How to prove $e^x\ge \left(1+\frac xn\right)^n$ for any real numbers $x, n > 0$ Can someone provide a detailed proof? I saw a proof here 

$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right) \tag 1\\\\
&\ge \left(1+\frac{-x}{n+x}\right)\left(1+\frac xn\right)\tag 2\\\\
&=1
\end{align}$$
where in going from (1) to (2) we used Bernoulli's Inequality. Note
  that (2) is valid whenever $n>−x$ or $x>−n$. Since $e_n(x)$
  monotonically increases and is bounded above by $e^x$, then $$e^x\ge
 \left(1+\frac xn\right)^n \tag 3$$ for all $n\ge 1$.

But I don't know how do we get $(1)$.

By the binimial theorem $$\left(1+\frac{x}{n}\right)^n=1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+...+\frac{n(n-1)...(n-n+1)}{n!}\cdot\frac{x^n}{n^n}=$$
$$=1+x+\left(1-\frac{1}{n}\right)\frac{x^2}{2!}+...+\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{n-1}{n}\right)\frac{x^n}{n!}<$$
$$<1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}<1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...=e^x.$$
