Show that $2^n>n$ for $n\in \mathbb{N}$ 
Show that $2^n>n$ for $n\in \mathbb{N}$

I can solve this by mathematical induction. Is there any other method to solve.
 A: $${ \left( 1+1 \right)  }^{ n }=1+n+\frac { n\left( n-1 \right)  }{ 2 } +..+1>n$$
A: By the arithmetic-geometric mean inequality,
$$\sqrt[n]n\le{n+1+1+\cdots+1\over n}={2n-1\over n}\lt 2$$
(where the first numerator has one $n$ and $n-1$ $1$'s). Of course this leaves the question of proving AGM without using induction.
A: With set theory:
Observe that $2^n$ is the cardinality of the power set of an $n$-element set. The power set of a finite set must have greater cardinality than that of the original set, because it contains all singletons, as well as the empty set, and thus has at least $n+1$ elements, which is trivially greater then $n$.
For $n>1$, the power set also contains other, larger subsets, but the proof is already complete.
A: Using the binomial expansion,
$$2^n = (1+1)^n = \sum_{i=0}^n {n\choose i}\geq 1+n$$
since every term in the sum is positive, and both ${n\choose 1}=n$ and ${n\choose 0}=1$ appear in the sum. 
A: With analysis:
$2^1 >1$, and for any $x>1$, $\;(2^x)'>(x)'=1$, since


*

*$(2^x)'=2^x\ln 2>2\ln 2=\ln4$

*$4>\mathrm e$.


Hence, by a standard corollary  of the Mean value theorem,  $\;2^x>x\;$ for all $\;x\ge 1$.
