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