Prove that $\left(1+\frac1 n\right)^n > 2$ I'm trying to demonstrate that $\left( 1+\frac1 n \right)^n$ is bigger than $2$. I have tried to prove that $\left( 1+\frac1 n \right)^n$ is smaller than $\left( 1+\frac1{n+1} \right)^{n+1}$ by expanding 
$\left( 1+\frac1n \right)^n = \sum\limits_{i=0}^n \left( \frac{n}{k} \right) \frac{1}{n^k}$ and $\left( 1+\frac1{n+1} \right)^{n+1} = \sum\limits_{i=0}^{n+1} \left( \frac{(n+1)}{k} \right) \frac{1}{(n+1)^k}$ but it doesn't seem to work.
What am I missing? Also, is there a method to demonstrate that without induction?
 A: You just need to consider the first $2$ terms in the binomial expansion
\begin{eqnarray*}
\left( 1+ \frac{1}{n} \right)^{n} = \sum_{i=0}^{n}\binom{n}{i}\frac{1}{n^i}=1+n \frac{1}{n} +\cdots \geq 2.
\end{eqnarray*}
A: Let $f(x) = (1+x)^n$. Note that $f$ is convex for $x \ge 0$ and so
$f(x) \ge f(0)+f'(0) x = 1+xn$. Hence $f({1 \over n}) \ge 2$.
A: \begin{align}
\left(1+\frac1n\right)^n &= \sum_{k=0}^n {n \choose k} \frac1{n^k} \\
&= \sum_{k=0}^n \frac{n(n-1)\cdots(n-k+1)}{k!n^k} \\
&= \sum_{k=0}^n \frac1{k!}\left(1-\frac1n\right)\left(1-\frac2n\right)\cdots \left(1-\frac{k-1}n\right)
\end{align}
so
\begin{align}
\left(1+\frac1{n+1}\right)^{n+1} &= \sum_{k=0}^{n+1} \frac1{k!}\left(1-\frac1{n+1}\right)\left(1-\frac2{n+1}\right)\cdots \left(1-\frac{k-1}{n+1}\right)\\
&> \sum_{k=0}^{n} \frac1{k!}\left(1-\frac1{n}\right)\left(1-\frac2{n}\right)\cdots \left(1-\frac{k-1}{n}\right)\\
&= \left(1+\frac1n\right)^n
\end{align}
On the other hand, for $n = 1$ we have
$$\left(1+\frac11\right)^1 = 2$$
so $\left(1+\frac1n\right)^n > 2, \forall n \ge 2$.
A: Another way is to prove first that your sequence is monotonically increasing like has been done here:
I have to show $(1+\frac1n)^n$ is monotonically increasing sequence
... and since your first term is $2$, it follows that the subsequent ones are larger than $2$.
