Any alternate proof for $2^n>n$? The normal approach for these kinds of problem is to use the mathematical induction and prove that $2^n>n$ for any natural number $n$
Case 1: $(n=1)$
$2^1=2>1$, thus the formula holds for $n=1$
Case 2: (let us assume that this statement holds for any arbitrary natural number $m$)
That implies $2^m>m$ for some natural number $m$.
Then,
$2^{m+1}=2^m \cdot 2>m \cdot 2 \geq m+1$
Thus as the statement holds for an arbitrary natural number $m$ implies that it holds for $m+1$ and thus by mathematical induction, it is proved that $2^n>n$ for any natural number $n$.
Is there any other ways to prove this problem?
I tried proving this by contradiction taking initially $2^n \leq n$, but couldn't go much far.
Any help or idea would be very much appreciated.
 A: We have $|\mathcal{P}(\{1,\dots,n \})| = 2^n$ and $|\{1,\dots, n\}| = n$ and there is an obvious non-surjective injection $$\{1, \dots, n \} \hookrightarrow \mathcal{P}(\{1,\dots,n \}).$$
Edit: Alternatively, $$2^n = (1+1)^n = \sum_{k=0}^n \binom{n}{k} \geq \sum_{k=0}^n 1 = n+1.$$ Note that in its spirit, this is not really a different proof. The binomial formula in this case is the same thing as counting subsets of a set.
A quick comment: Heuristically, because this inequality is so weak, there will be plenty of ad hoc arguments.
A: $$n=\underbrace{1+1+\cdots+1}_{n\ \text{terms}}$$
$$2^n=\underbrace{2+2+2^2+2^3+\cdots+2^{n-1}}_{n\ \text{terms}}$$
$2^n>n$ follows from the fact that $2^i>1$ for $i\ge1$.
A: By a combinatoric argument, $2^n$ is the number of subsets of a set with $n$ elements which is always greater than the number of elements.
Refer to the related

*

*The total number of subsets is $2^n$ for $n$ elements

As an alternative, by derivative, let define
$$f(x)=2^x-x  \implies f'(x) =2^x \log 2-1$$
with $f(1)=1$ and $f'(1)>0$ therefore $\forall x\ge 1$
$$2^x-x \ge 0 \iff 2^x\ge x$$
