By using mathematical induction prove that (n+1)! > 2^(n+1) for n, where n is a positive integer greater than or equal to 4

The result is trivially true for $k=4$ since, $(4+1)!=120>32=2^{5}.$ Lets us assume it to be true for $k=n.$ We prove for $k=n+1.$ Now,

$(n+2)!=(n+1)!(n+2)>2^{n+1}(n+2)$ by induction hypothesis. Now $n>4\Rightarrow n+2>6 >2.$ Hence we get, $2^{n+1}(n+ 2)>2^{n+1}\times 2=2^{n+2}.$ So we are done by induction.