Show that $e^x > \bigg(1+ \frac xn\bigg)^n $ Show that $$e^x > \bigg(1+ \frac xn\bigg)^n$$
I have thought about logarithmically deriving the sides but I cannot achieve the desired results. 
Is that the correct way to go with this question?
 A: There's something much simpler you can do to both sides, which enables you to express the problem in terms of $x/n$ rather than $x$.
A: $$\left(\left(1+\frac{x}{n}\right)^n\right)_{n\in \mathbb N}$$ 
is an increasing sequence that converge to $e^x$, the result follow.
A: Let $x\in {\mathbb R}_{\ge 0}$. Now, note that 
$$
(1+\frac{x}{n})^n =
\sum_{k=0}^n \frac{n!}{k!(n-k)!}\frac{x^k}{n^k}=
\sum_{k=0}^n \frac{n!}{(n-k)!n^k}\frac{x^k}{k!}.$$
While this is equivalent with
$$
\sum_{k=0}^n \frac{n(n-1)(n-2)...(n-k+1)}{n^k}\frac{x^k}{k!}=
\sum_{k=0}^n (1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{k-1}{n})\frac{x^k}{k!}. 
$$
Thus
$$
(1+\frac{x}{n})^n = \sum_{k=0}^n (1-\frac{1}{n})(1-\frac{2}{n})...(1-\frac{k-1}{n})\frac{x^k}{k!} \le
\sum_{k=0}^n \frac{x^k}{k!} \le \sum_{k=0}^\infty \frac{x^k}{k!} =e^x.
$$
For $x\in {\mathbb R}_{< 0}$, one can develop similar trend.
A: Raise your inequality to the power $\frac{1}{n}$ to obtain, equivalently, that
$$
e^{\frac{x}{n}}>1+\frac{x}{n}.
$$
Next observe, for example using the first derivative, that $0$ is a global minimum for the function $f(y)=e^y-y-1$ ($y\in\mathbb{R}$). Deduce that $f(y)\ge f(0)=0$ for every $y$ (hence also for $y=\frac{x}{n}$) with equality iff $y=0$ ($x=0$).
A: Hint: $$\ln \Big(1 + \frac{x}{n}\Big) \leq \frac{x}{n} $$
For $\frac{x}{n} > -1$.
Alternatively 
$f: I \to \mathbb R$ defined as $f(x) = e^x$ is convex   ($f''(x) > 0 ,\forall x \in I$), then for $x \in I = (-\infty, \infty)$
$$f(x) \geq f(0) + f'(0)(x - 0) \implies e^x \geq 1 + x$$
