If \(n\) is an integer, you can use induction. First show the inequality holds for \(n=1\). You have \(2^1=2>1\), which is true.
Now assume this holds in general for \(n=k\), i.e. that \(2^k>k\). We want to prove the statement then must hold for \(n=k+1\).
Because \(2^k>k\), you have
\(2^{k+1}=2\times2^k>2k\)
and this must be greater than \(k+1\) for the statement to be true, so we require
\(2k>k+1\)
for \(k>1\). Well this is obviously true, because solving the inequality gives \(3k>1\implies k>\dfrac13\). So you're done.
If you \(n\) is any real number, you can use derivatives to show that \(2^n\) increases monotonically and faster than \(n\).
