Answer
The inequality was proved for $n=1$
The inequality is correct if $n$ is changed by $n+1$
Work Step by Step
Let's prove the inequality for $n=4$:
$4!\gt2^4$
$4\times3\times2\times1\gt16$
$24\gt16$
It is correct!
Now, suppose that the inequality is correct, that is:
$n!\gt2^n$ for $n\geq4$
Now, let's prove the inequality for $n+1$:
$(n+1)!=(n+1)n(n-1)(n-2)...3\times2\times1=(n+1)n!$
Since $n\geq4$ we have that $n+1\geq5\gt2$:
$(n+1)!=(n+1)n!\gt2(n!)\gt2(2^n)=2^{n+1}$
$(n+1)!\gt2^{n+1}$
That is exactly the given inequality if $n$ is changed by $n+1$
