show $(1+\frac{x}{n})^n \leq e^x$ I want to show that $(1+\frac{x}{n})^n \leq e^x$ for all $x \geq -n$, $n\in \mathbb{N}$.
I already showed that positive case ($x>0$) and zero case $(x=0)$.
Not sure how to approach the negative case, because in the case $x>0$ I used that $x^n$ is an increasing function on $(0, \infty)$. But I can't do the same when $x$ is negative.
 A: You first want to prove $1 + x \le e^x$ for all $x$, which holds because $e^x$ is convex and $1+x$ is the tangent line to $e^x$ at $(0,1)$, for example.
From $x \ge -n$ we get $\frac xn \ge -1$, or $1 + \frac xn \ge 0$. By the inequality, $1 + \frac xn \le e^{x/n}$. Because $t \mapsto t^n$ is an increasing function on $[0,\infty)$ and both sides are positive, $(1 + \frac xn)^n \le (e^{x/n})^n = e^x$.
A: Assuming you know that
\begin{align}
\lim_{n\rightarrow \infty}\left( 1+ \frac{x}{n}\right)^n = e^x,
\end{align}
then it suffices to prove that
\begin{align}
\left( 1+ \frac{x}{n}\right)^n \le \left( 1+ \frac{x}{n+1}\right)^{n+1}
\end{align}
for all $n \in \mathbb{N}$.
By Bernoulli's inequality, we have that
\begin{align}
\frac{\left( 1+ \frac{x}{n+1}\right)^{n+1}}{\left( 1+ \frac{x}{n}\right)^{n+1}} =&\ \left( 1+\frac{-\frac{x}{n(n+1)}}{1+\frac{x}{n}}\right)^{n+1} = \left( 1-\frac{x}{n(n+1)+(n+1)x}\right)^{n+1} \\
 \ge&\ 1-(n+1)\left(\frac{x}{n(n+1)+(n+1)x} \right) = 1-\frac{x}{n+x} = \frac{1}{1+\frac{x}{n}}.
\end{align}
Multiplying both side by $(1+x/n)^{n+1}$ yields the desired inequality.
Remark: This is a standard proof given in a first course of analysis without assuming any knowledge of derivatives.
