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 $(1)$.
 A: Note that we have
$$\begin{align}
\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}n\right)^n}&=\left(\frac{1+\frac{x}{n+1}}{1+\frac xn}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(\frac{\frac{n+1+x}{n+1}}{\frac {n+x}n}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(\frac{n(n+1+x)}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(1+\frac{n(n+1+x)-(n+1)(n+x)}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
&=\left(1+\frac{-x}{(n+1)(n+x)}\right)^{n+1}\left(1+\frac{x}n\right)\\\\
\end{align}$$
as was to be shown!
A: By the binimial theorem $$\left(1+\frac{x}{n}\right)^n=1+n\cdot\frac{x}{n}+\frac{n(n-1)}{2!}\cdot\frac{x^2}{n^2}+...+\frac{n(n-1)...(n-n+1)}{n!}\cdot\frac{x^n}{n^n}=$$
$$=1+x+\left(1-\frac{1}{n}\right)\frac{x^2}{2!}+...+\left(1-\frac{1}{n}\right)\left(1-\frac{2}{n}\right)...\left(1-\frac{n-1}{n}\right)\frac{x^n}{n!}<$$
$$<1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}<1+x+\frac{x^2}{2!}+...+\frac{x^n}{n!}+...=e^x.$$
A: \begin{align}
\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n} &= \frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^{n + 1}}\left(1+\frac xn\right) \\
&= \frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left( \frac{n + x}{n} \right)^{n + 1}}\left(1+\frac xn\right) \\
&= \left(1+\frac x{n+1}\right)^{n+1}\left( \frac{n}{n + x} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left(\left(1+\frac x{n+1}\right) \frac{n}{n + x} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n}{n + x} + \frac{nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n(n+1)}{(n+1)(n + x)} + \frac{nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n(n + 1) + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{n(n + 1) + [x(n+1) - x(n+1)] + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( \frac{(n + x)(n + 1) - x(n+1) + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( 1 + \frac{- x(n+1) + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( 1 + \frac{- xn - x + nx}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
&= \left( 1 + \frac{-x}{(n + 1)(n + x)} \right)^{n + 1}\left(1+\frac xn\right) \\
\end{align}
A: An elementary inequality in Calculus is $\ln (1+x) \leq x$ for $x>0$. (You can prove this by noting that the derivative of $x-\ln (1+x)$ is positive, so its minimum ia attained at $x=0$). Now $\ln (1+x/n) \leq x/n$ so $n \ln (1+x/n) \leq x$. Take exponential on both sides to get $(e^{\ln (1+x/n)})^{n} \leq e^{x}$ or $(1+x/n)^{n} \leq e^{x}$.
