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.

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 $\forall x\ge 1$
$$2^x-x \ge 0 \iff 2^x\ge x$$
